Quadratic Equation
A quadratic equation is an equation of the form
\( ax^2+bx+c=0\)
where the leading coefficient \(a \ne 0\), x represents an unknown, and a, b, and c represent known numbers.
If a = 0, then the equation is linear, not quadratic, as there is no \(ax^2 \)term.
Some key concepts of Quadratic Equation.
 The standard form of Quadratic Equation is
\( ax^2 + bx + c = 0\) where \(a \ne 0\)
 Because the quadratic equation involves only one unknown, it is called "univariate".
 The numbers a, b, and c are the coefficients, they are also respectively called quadratic coefficient, linear coefficient, and the constant or free term.
 The quadratic equation contains only powers of x that are nonnegative integers, and therefore it is a polynomial equation. In particular, it is a seconddegree polynomial equation, since the greatest power is two.
 Equations such as \( x^2 = 64, x^2 5x = 0\), and \(x^2 + 4x = 5\) are examples of quadratic equations. This is because in each of these equations, the greatest exponent of the variable x is 2.
 A quadratic equation can be factored into an equivalent equation
\( ax^2+bx+c=(x\alpha)(x\beta)=0\)
where \(\alpha\) and \(\beta\) are the solutions for x.
 The values of x that satisfy the equation are called solutions, roots or zeros of Quadratic Equation.
A quadratic equation has at most two solutions. If there is only one solution, it is a double root.
 A quadratic equation always has two roots, if complex roots are included and a double root is counted for two.
 A quadratic polynomial has a parabola as its graph
We can obtain three different forms of a quadratic function
 The Standard form representation of Quadratic Equation is : \(f(x) = ax^2 + bx + c\).
Here, function f(x) has two zeros, a double zero, or no zero according to whether its discriminant \(b^24ac\) is positive, zero, or negative, respectively
 The Vertex form representation of Quadratic Equation is : \( f (x) = a(xp)^2 + q\). Here, the point \((p, q)\) is called the vertex of the Quadratic Equation.
The graph of f(x) has bilateral symmetry with respect to the vertical line defined by \(x = p\).
 If a > 0, f(x) is
decreasing on (−∞, p] and therefore increasing on [p,∞).
 If a < 0, then f(x) is increasing on (−∞, p] and therefore decreasing on [p,∞).
 If a > 0, then f(x) achieves its minimum at p.
 If a < 0, then f(x) achieves its maximum at p.
 The Factored form representation of Quadratic Equation is: \(f (x) = a(x−\alpha)(x−\beta)\), where \(\alpha, \beta\) are the zeros of f(x). Also \(\alpha+ \beta=\frac{b}{a}\) and \(\alpha \beta=\frac{c}{a}\)
Each of the three representations of a Quadratic Equation reveals a different facet of the function \(f(x) = ax^2 + bx + c\). The quadratic formula is expressed in terms standard notion, the vertex form displays the line of symmetry of the graph of f(x) and also where it achieves its maximum or minimum, and factored form displays its zeros explicitly. Together, the three representations give a wellrounded picture of Quadratic Equation.
Solution of Quaadratic Equation
A quadratic equation is an equation \( ax^2+bx+c=0\) can be solved in three different ways:
 Graphical Method
 Factoring Method
 Formula Method
Not every quadratic equation can be solved by factoring of graphing method. In this case, we need to use the quadratic formula.
The Quadratic Formula is
\( x= \frac{b \pm \sqrt {b^24ac}}{2a} \)
Graphical Method
A solution of a quadratic equation is also called a root of the equation. The intuitive meaning of the root of a quadratic equation
can be given pictorially, as follows.
The graph of y = 2x^{2} − x − 3 intersects the xaxis at (−1, 0) and (1.5, 0), and a simple computation confirms that −1 and 1.5 are roots of 2x^{2} − x − 3 = 0
Prove that two roots of quadratic equations \(ax^2 + bx + c = 0\) are \(\frac{b  \sqrt{b^24ac}}{2a}\) and \(\frac{b + \sqrt{b^24ac}}{2a}\)
Proof
Given quadratic equation is
\(ax^2 + bx + c = 0\)
or \(ax^2 + bx =c\)
Dividing both sides by a, we get
\(x^2 + \frac{b}{a}x =  \frac{c}{a}\)
Adding \(\frac{b^2}{4a^2}\) on both sides we get
\(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}= \frac{b^2}{4a^2} \frac{c}{a}\)
or \(\left( x+\frac{b}{2a} \right )^2=\frac{b^24ac}{4a^2}\)
Taking square roots we get
\(x+\frac{b}{2a}=\frac{\pm \sqrt{b^24ac}}{2a}\)
or \(x=\frac{b\pm \sqrt{b^24ac}}{2a}\)
So, the roots of quadratic equations \(ax^2 + bx + c = 0\) are \(\frac{b  \sqrt{b^24ac}}{2a}\) and \(\frac{b + \sqrt{b^24ac}}{2a}\)
To solve a Quadratic equation using the formula, we use the following steps:
 Put the quadratic equation into standard form : \(ax^2 + bx + c = 0\)
 Write out the value for a, b, and c
 Substitute value in the formula, solve and get the roots
 Check each root in the original equation
Nature of Roots
In the quadratic equation \(ax^2+bx+c=0\), the expression \(b^24ac\) is called the discriminant, and is often represented using an upper case D or an upper case Greek delta \( \Delta \) given as
\( \Delta=b^24ac \)
A quadratic equation with real coefficients can have either one or two distinct real roots, or two distinct complex roots. In this case, the discriminant determines the number and nature of the roots. The figure below shows its different cases.



Δ is positive 
Δ is zero 
Δ is negative 
The figure given above plots three quadratic equations to illustrate the effects of discriminant values. There are three cases:
 When the discriminant \( \Delta\) is positive, the curve of quadratic equation (parabola) intersects the xaxis at two points. Then there are two distinct roots
\( \frac {b+\sqrt{ \Delta }}{2a}\) and \( \frac {b\sqrt{ \Delta }}{2a}\)
both of which are real numbers.
For quadratic equations with rational coefficients,
 if the discriminant is a square number, then the roots are rational
 if the discriminant is a NOT square number, then the roots are irrational
This kinds of irrational roots always occur in conjugate pairs, which are of the form
\( \alpha= \frac {b+\sqrt{ b^24ac }}{2a}\) and \( \beta= \frac {b\sqrt{ b^24ac }}{2a}\)
or \( \alpha= \frac {b}{2a}+\frac{\sqrt{ b^24ac }}{2a}\) and \( \beta= \frac {b}{2a}\frac{\sqrt{ b^24ac }}{2a}\)
or \( \alpha= p+\sqrt{q}\) and \( \beta= p\sqrt{q}\)
 When the discriminant \( \Delta\) is zero, the vertex of the curve of quadratic equation (parabola) touches the xaxis at a single point. Then there is exactly one real root
\( \frac{b}{2a}\)
sometimes called a repeated or double root.
 When the discriminant \( \Delta\) is negative, the curve of quadratic equation (parabola) does not intersect the xaxis at all. Then there are no real roots. Rather, there are two distinct (nonreal) complex roots
This kinds of complex roots always occur in conjugate pairs, which are of the form
\( \alpha= \frac {b+\sqrt{ b^24ac }}{2a}\) and \( \beta= \frac {b\sqrt{ b^24ac }}{2a}\)
or \( \alpha= \frac {b}{2a}+\frac{\sqrt{ b^24ac }}{2a}\) and \( \beta= \frac {b}{2a}\frac{\sqrt{ b^24ac }}{2a}\)
or \( \alpha= p+i q\) and \( \beta= piq\)
In these expressions i is the imaginary unit.
 Thus the roots are distinct if and only if the discriminant is nonzero, and the roots are real if and only if the discriminant is nonnegative.
Based on the discriminant \( b^24ac\) of the quadratic equation \( ax^2+bx+c=0\) , the corresponding solution will be
If the discriminant is 
then there (roots) 
positive \( b^24ac >0\) and a perfect square 
are two rational roots 
positive \( b^24ac >0\) and NOT a perfect square 
are two irrational roots 
zero: \( b^24ac=0\) 
is one rational roots 
negative: \( b^24ac < 0\) 
are two complex roots 
Graph of Quadratic Equation
The graph of a quadratic equation is a Ushaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry.
 If we want to identify the vertex of the parabola form of a quadratic function, then we convert \(f(x)=ax^2+bx+c\) in the form of \(f(x)=a(xh)^2+k\), where (h,k) is the vertex.
 the hcoordinate of the vertex is given by \(h=\frac{b}{2a}\)
 the kcoordinate of the vertex is given by \(k=f(h)= f\left (\frac{b}{2a} \right )\)
 the line of symmetry is given by \(x=h\)
Given a quadratic equation \(ax^2+bx+c=0\), based on the value of \(a\) and \(b^24ac\), there are six possibilities in the graph.
a>0 



Δ is positive
\(x^25x+4=0\) 
Δ is zero
\(x^26x+9=0\) 
Δ is negative
\(x^24x+5=0\) 
a<0 



Δ is positive
\(x^25x4=0\) 
Δ is zero
\(x^26x9=0\) 
Δ is negative
\(x^24x5=0\) 
Note: the example of the above graph are given below.
graph of a quadratic function 
\(b^24ac > 0\) 
\(b^24ac=0\) 
\(b^24ac < 0 \) 
\(a> 0\) 
Case 1 
Case 2 
Case 3 
\(a < 0\) 
Case 4 
Case 5 
Case 6 
case 1: \(a > 0 ,b^2 – 4ac > 0\)
Case 1: When a > 0 and b2 – 4ac > 0
The graph of a quadratic equation will be concave upwards and will intersect the xaxis at two points α and β with α < β. The quadratic equation will have two real roots (α and β), and the curve will always lie above the xaxis.
 The quadratic function f(x) will be positive, i.e., f(x) > 0, for the values of x lying in the interval (∞, α) ∪ (β, ∞)
 The quadratic function f(x) will be equal to zero, i.e., f(x) = 0, if x = α or β
 The quadratic function f(x) will be negative, i.e., f(x) < 0, for the values of x lying in the interval (α, β)
case 2: \(a> 0 ,b^2 – 4ac = 0\)
Case 2: When a > 0 and b2 – 4ac = 0
The graph of a quadratic equation will be concave upwards and will touch the xaxis at a point b/2a. The quadratic equation will have two equal real roots, i.e., α = β. The quadratic function f(x) will be positive, i.e., 0 ≤ f(x), x ∈ R.
case 3: \(a> 0 ,b^2 – 4ac = 0\)
Case 3: When a > 0 and b2 – 4ac < 0
The graph of a quadratic equation will be concave upwards and will not intersect the xaxis. The quadratic equation will have imaginary roots, and the curve will always lie above the xaxis. The quadratic function f(x) will be positive, i.e., f(x) > 0, x ∈ R.
case 4: \(a< 0 ,b^2 – 4ac > 0\)
Case 4: When a < 0 and b2 – 4ac > 0
The graph of a quadratic equation will be concave downwards and will intersect the xaxis at two points α and β with α < β. The quadratic equation will have two real roots (α and β), and the curve will always lie below the xaxis.
 The quadratic function f(x) will be negative, i.e., f(x) > 0, for the values of x lying in the interval (∞, α) ∪ (β, ∞)
 The quadratic function f(x) will be equal to zero, i.e., f(x) = 0, if x = α or β
 The quadratic function f(x) will be positive, i.e., f(x) < 0, for the values of x lying in the interval (α, β)
case 5: \(a< 0 ,b^2 – 4ac = 0\)
Case 5: When a < 0 and b2 – 4ac = 0
The graph of a quadratic equation will be concave downwards and will touch the xaxis at a point b/2a. The quadratic equation will have two equal real roots, i.e., α = β. The quadratic function f(x) will be negative, i.e., 0 ≥ f(x), x ∈ R.
case 6: \(a< 0 ,b^2 – 4ac > 0\)
Case 6: When a < 0 and b2 – 4ac < 0
The graph of a quadratic equation will be concave downwards and will not intersect the xaxis. The quadratic equation will have imaginary roots, and the curve will always lie below the xaxis. The quadratic function f(x) will be negative, i.e., f(x) < 0, x ∈ R.
Relation between Roots and Cofficients
Given a quadratic equation \(ax^2 + bx + c = 0\), we know that its roots are given by
\( \alpha=\frac{b  \sqrt{b^24ac}}{2a}\) and \(\beta =\frac{b + \sqrt{b^24ac}}{2a}\)
By addition, we get that
\( \alpha+\beta=\frac{b  \sqrt{b^24ac}}{2a}+\frac{b + \sqrt{b^24ac}}{2a}\)
or\( \alpha+\beta=\frac{2b}{2a}\)
or\( \alpha+\beta=\frac{b}{a}\)
Again, By multiplication, we get that
\( \alpha. \beta=\frac{b  \sqrt{b^24ac}}{2a}. \frac{b + \sqrt{b^24ac}}{2a}\)
or\( \alpha\beta=\frac{(b)^2(b^24ac)^2}{4a^2}\)
or\( \alpha \beta=\frac{c}{a}\)
Some special cases
 In a quadratic equation \(ax^2 + bx + c = 0\), if \(b=0\) then
\( \alpha+\beta=\frac{b}{a}\)
or \( \alpha+\beta=\frac{0}{a}\)
or \( \alpha+\beta=0\)
or \( \alpha=\beta\)
This means, the two roots are equal but opposite in sign.
For example: \(ax^2 + c = 0\).
 In a quadratic equation \(ax^2 + bx + c = 0\), if \(c=0\) then
\( \alpha \beta=\frac{c}{a}\)
or \( \alpha \beta=\frac{0}{a}\)
or \( \alpha \beta=0\)
or \( \alpha =0\) or \( \beta =0\)
This means, the one of the two roots must be zero.
For example: \(ax^2 + bx = 0\).
 In a quadratic equation \(ax^2 + bx + c = 0\), if \(b=0,c=0\) then
\( \alpha=\beta\) and \( \alpha =0\) or \( \beta =0\)
or \( \alpha=0\) and \( \beta =0\)
This means, the both roots are zero.
For example: \(ax^2 = 0\).
If α and β are the zeros of a quadratic equation \(f(x):ax^2+bx+c=0\) then the function equation can be written as
\(ax^2+bx+c=0 \)
or \( x^2+ \frac{b}{a} x+\frac{c}{a}=0 \)
or \( x^2  \left ( \frac{b}{a} \right ) x+ \left ( \frac{c}{a} \right ) =0 \)
or \( x^2−(\alpha+\beta)x+(\alpha.\beta)=0\)
or \(x^2− (\text{ sum of roots }) x+ (\text{ product of roots}) =0\)
Symmetric Function
Let α and β be the roots of the quadratic equation \(ax^2 + bx + c = 0\), (a ≠ 0), then the expressions of the form
\( \alpha + \beta , \alpha \beta , \alpha ^2 + \beta ^2 , \alpha ^2  \beta ^2 , \frac{1}{\alpha ^2} + \frac{1}{\beta ^2}\) etc., which can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\) , are known as symmetric functions of the roots α and β.
So, a function is symmetric in terms of α and β, either the relation remains the same or is multiplied by 1. Hence, if one of the following is satisfied, then we say function is symmetric of for α and β
 an expression in α and β which remains same when α and β are interchanged
 an expression in α and β which remains same when multiplied by 1, when α and β are interchanged
Thus
\( \alpha + \beta\) is a symmetric function
Also
\( \alpha  \beta\) is a symmetric function because while interchanging α and β then \( \alpha  \beta= \sqrt{ (\alpha + \beta)^2 4 \alpha \beta }\) or \( \alpha  \beta= (\beta  \alpha) \)
While
3α +2β is NOT a symmetric function because while interchanging α and β then 3α +2β≠ 3β + 2α
Some more example of symmetric functions are given below
 \( \alpha + \beta\)
 \( \alpha \beta\)
 \( \alpha  \beta\)
because \( \alpha  \beta =\sqrt{(\alpha + \beta)^24\alpha \beta }\)
means \( \alpha  \beta\) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
 \( \alpha ^2 + \beta ^2\)
because \( \alpha ^2 + \beta ^2 =(\alpha + \beta)^22\alpha \beta \)
means \( \alpha ^2 + \beta ^2\) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
 \( \alpha ^2  \beta ^2\)
because \( \alpha ^2  \beta ^2 = (\alpha + \beta) \sqrt{(\alpha + \beta)^24\alpha \beta }\)
means \( \alpha ^2  \beta ^2 \) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
 \( (\alpha  \beta )^2\)
because \( (\alpha  \beta )^2 =(\alpha + \beta)^24\alpha \beta \)
means \( (\alpha  \beta )^2 \) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
 \( \alpha ^3 + \beta ^3\)
because \( \alpha ^3 + \beta ^3 =(\alpha + \beta)^33\alpha \beta (\alpha + \beta)\)
means \( \alpha ^3 + \beta ^3\) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
 \( \alpha ^3  \beta ^3\)
because \( \alpha ^3  \beta ^3 =\sqrt{(\alpha + \beta)^24\alpha \beta } ( \alpha ^2 + \alpha \beta + \beta ^2) \)
means \( \alpha ^3  \beta ^3 \) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
 \( (\alpha  \beta )^3\)
because \( (\alpha  \beta )^3 = \left ( \sqrt{(\alpha + \beta)^24\alpha \beta } \right )^3\)
means \( (\alpha  \beta )^3 \) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
 \( \alpha ^4  \beta ^4\)
because \( \alpha ^4  \beta ^4 = (\alpha ^ 2+ \beta ^2)(\alpha + \beta) \sqrt{(\alpha + \beta)^24\alpha \beta }\)
means \( \alpha ^4  \beta ^4 \) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
 \( \frac{1}{\beta }+ \frac{1}{\alpha }\)
because \( \frac{1}{\beta }+ \frac{1}{\alpha } = \frac{\alpha + \beta}{\alpha \beta}\)
means \( \frac{1}{\beta }+ \frac{1}{\alpha } \) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
 \(\frac{\alpha}{\beta }+ \frac{\beta}{\alpha } \)
because \( \frac{\alpha}{\beta }+ \frac{\beta}{\alpha } =\frac{\alpha ^2 + \beta ^2}{\alpha \beta}\)
means \( \frac{\alpha}{\beta }+ \frac{\beta}{\alpha }\) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
 \(\frac{\alpha ^2}{\beta }+ \frac{\beta ^2}{\alpha } \)
because \( \frac{\alpha ^2}{\beta }+ \frac{\beta ^2}{\alpha }=\frac{\alpha ^3 + \beta ^3}{\alpha \beta}\)
means \( \frac{\alpha ^2}{\beta }+ \frac{\beta ^2}{\alpha }\) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
Example to workout 👉 Click Here
 If α and β are the roots of the quadratic \(3x^2 4x 1 = 0\), determine the values of the following expressions.
 \( \alpha + \beta\)
 \( \alpha \beta\)
 \( \alpha ^2 + \beta^2\)
 \( \alpha ^3 + \beta^3\)
 \( \frac{1}{\beta }+ \frac{1}{\alpha }\)
 \(\frac{\alpha}{\beta }+ \frac{\beta}{\alpha } \)
 \(\frac{\alpha ^2}{\beta }+ \frac{\beta ^2}{\alpha } \)
 If α and β are the roots of the quadratic \(ax^2 + bx + c = 0\), (a ≠ 0), determine the values of the following expressions in terms of a, b and c.
 \( \alpha + \beta\)
 \( \alpha \beta\)
 \( \alpha ^2 + \beta^2\)
 \( \alpha ^3 + \beta^3\)
 \( \frac{1}{\beta }+ \frac{1}{\alpha }\)
 \(\frac{\alpha}{\beta }+ \frac{\beta}{\alpha } \)
 \(\frac{\alpha ^2}{\beta }+ \frac{\beta ^2}{\alpha } \)
Common Root

One root common: Let \(\alpha\) be the common root of two quadratic equations \(a_1x^2+b_1x+c_1=0\) and \(a_2x^2+b_2x+c_2=0\) then
\(a_1 \alpha ^2+b_1 \alpha +c_1=0\) and \(a_2 \alpha ^2+b_2 \alpha +c_2=0\)
By crammer's rule, we can write
\( \frac{\alpha ^2}{ \begin{vmatrix} c_1 & b_1\\ c_2 & b_2
\end{vmatrix} } = \frac{\alpha}{ \begin{vmatrix} a_1 & c_1\\ a_2 & c_2
\end{vmatrix} } = \frac{1}{ \begin{vmatrix} a_1 & b_1\\ a_2 & b_2
\end{vmatrix} } \)
or\( \frac{ \alpha ^2}{b_1c_2b_2c_1} =\frac{ \alpha}{a_2c_1a_1c_2} =\frac{1}{a_1b_2a_2b_1}\)
or \(\alpha= \frac{a_2c_1a_1c_2}{a_1b_2a_2b_1} =\frac{b_1c_2b_2c_1}{a_2c_1a_1c_2} \)
or \( (a_2c_1a_1c_2)^2 = (a_1b_2a_2b_1) (b_1c_2b_2c_1) \)
Thus, the condition having a common root \(\alpha\) in two quadratic equations \(a_1x^2+b_1x+c_1=0\) and \(a_2x^2+b_2x+c_2=0\) is
\( (a_2c_1a_1c_2)^2 = (a_1b_2a_2b_1) (b_1c_2b_2c_1) \)

Both roots common:Let α, β be the common roots of the quadratic equations \(a_1x^2+b_1x+c_1=0\) and \(a_2x^2+b_2x+c_2=0\) then
Considering α, β as roots of \(a_1x^2+b_1x+c_1=0\), we get
\( \alpha + \beta =  \frac{b_1}{a_1},\alpha \beta = \frac{c_1}{a_1} \)
Again
Considering α, β as roots of \(a_2x^2+b_2x+c_2=0\), we get
\( \alpha + \beta =  \frac{b_2}{a_2},\alpha \beta = \frac{c_2}{a_2} \)
Therefore
\( \frac{b_1}{a_1}= \frac{b_2}{a_2}\) and \(\frac{c_1}{a_1}=\frac{c_2}{a_2} \)
or \( \frac{a_1}{a_1}= \frac{b_1}{b_2}\) and \(\frac{a_1}{a_2}=\frac{c_1}{c_2} \)
or \( \frac{a_1}{a_1}= \frac{b_1}{b_2}=\frac{c_1}{c_2} \)
This is the required condition.
Grade 11 Mathematics: Quadratic Equation [BCB Exercise 6.1]
 Determine the nature of the roots of each of the following equations
 \( x^26x+5=0 \)
Solution 👉 Click Here
Solution :1a
Given quadratic eqiation is \( x^26x+5=0 \)
Comparing the equation with \(ax^2+bx+c=0\), we get
\(a=1,b=6,c=5\)
Now, the discriminant of the equation is
\(b^24ac\) 
=\((6)^24(1)(5)\) 

\(=3620\) 

=16 
It shows that, the discriminant is
positive and perfect square, therefore, the nature of the roots is
two rational roots
 \( x^24x3=0 \)
Solution 👉 Click Here
Solution :1b
Given quadratic eqiation is \( x^24x3=0 \)
Comparing the equation with \(ax^2+bx+c=0\), we get
\(a=1,b=4,c=3\)
Now, the discriminant of the equation is
\(b^24ac\) 
=\((4)^24(1)(3)\) 

\(=16+12\) 

=28 
It shows that, the discriminant is
positive but NOT perfect square, therefore, the nature of the roots is
two irrational roots
 \( x^26x+9=0 \)
Solution 👉 Click Here
Solution :1c
Given quadratic eqiation is \( x^26x+9=0 \)
Comparing the equation with \(ax^2+bx+c=0\), we get
\(a=1,b=6,c=9\)
Now, the discriminant of the equation is
\(b^24ac\) 
=\((6)^24(1)(9)\) 

\(=3636\) 

=0 
It shows that, the discriminant is
zero, therefore, the nature of the roots is
two EQUAL rational roots
 \( 4x^24x+1=0 \)
Solution 👉 Click Here
Solution :1d
Given quadratic eqiation is \( 4x^24x+1=0 \)
Comparing the equation with \(ax^2+bx+c=0\), we get
\(a=4,b=4,c=1\)
Now, the discriminant of the equation is
\(b^24ac\) 
=\((4)^24(4)(1)\) 

\(=1616\) 

=0 
It shows that, the discriminant is
zero, therefore, the nature of the roots is
two EQUAL rational roots
 \( 2x^29x+35=0 \)
Solution 👉 Click Here
Solution :1e
Given quadratic eqiation is\( 2x^29x+35=0 \)
Comparing the equation with \(ax^2+bx+c=0\), we get
\(a=2,b=9,c=35\)
Now, the discriminant of the equation is
\(b^24ac\) 
=\((9)^24(2)(35)\) 

\(=81280\) 

=199 
It shows that, the discriminant is
Negative, therefore, the nature of the roots is
two complex numbers
 \( 4x^2+8x5=0 \)
Solution 👉 Click Here
Solution :1f
Given quadratic eqiation is \( 4x^2+8x5=0 \)
Comparing the equation with \(ax^2+bx+c=0\), we get
\(a=4,b=8,c=5\)
Now, the discriminant of the equation is
\(b^24ac\) 
=\((8)^24(4)(5)\) 

\(=64+80\) 

=144 
It shows that, the discriminant is
positive and perfect square, therefore, the nature of the roots is
two rational roots
 For what values of p will the equation \( 5x^2px+45=0\)?
Solution 👉 Click Here
Solution :2
 if the equation \( x^2+2(k+2)x+9k=0\) has equal roots, find k.
Solution 👉 Click Here
Solution :3
 For what value of a will the equation \(x^2(3a1)x+2(a^21)=0 \) have equal roots?
Solution 👉 Click Here
Solution :4
 If the roots of the equation \((a^2+b^2)x^22(ac+bd)x+(c^2+d^2)=0 \) are equal, then show that \(\frac{a}{b}=\frac{c}{d}\)
Solution 👉 Click Here
Solution :5
 Show that the roots of the equation \( (a^2bc)x^2+2(b^2ca)x+(c^2ab)=0\) will be equal, if either \(b=0\) or \(a^3+b^3+c^33abc=0\)
Solution 👉 Click Here
Solution :6
 If \(a,b,c\) are rational and \( a+b+c=0\), show that the roots \((b+ca)x^2+(c+ab)x+(a+bc)=0\) are rational.
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Solution :7
 Prove that the roots of the equation \((xa)(xb)=k^2\) are real for all values of k.
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Solution :8
 Show that the roots of the equation \(x^24abx+(a^2+2b^2)^2=0 \) are imaginary.
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Solution :9
 If the roots of the quadratic equation \(qx^2+2px+2q=0\) are real and unequal, prove that the roots of the equation \((p+q)x^2+2qx+(pq)=0\) are imaginary
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Solution :10
Grade 11 Mathematics: Quadratic Equation [BCB Exercise 6.2]
 From the equation whose roots are
 3,2
Solution 👉 Click Here
Solution :1a
Given roots roots are 3,2, thus
\( \text {sum of roots}=(3)+(2)=1, \text {product of roots}=(3)(2)=6\)
According to the question
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or \( x^2 \left ( 1 \right) x + \left ( 6 \right )=0\)
or \(x^2x 6=0\)
This is the required quadratic equation
 5,4
Solution 👉 Click Here
Solution :1b
Given roots roots are 5,4, thus
\( \text {sum of roots}=(5)+(4)=1, \text {product of roots}=(5)(4)=20\)
According to the question
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or \( x^2 \left ( 1 \right) x + \left ( 20 \right )=0\)
or \(x^2+x 20=0\)
This is the required quadratic equation
 \(\sqrt{3},\sqrt{3}\)
Solution 👉 Click Here
Solution :1c
Given roots roots are \(\sqrt{3},\sqrt{3}\), thus
\( \text {sum of roots}=(\sqrt{3})+(\sqrt{3})=0, \text {product of roots}=(\sqrt{3})(\sqrt{3})=3\)
According to the question
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or \( x^2 \left ( 0 \right) x + \left ( 3 \right )=0\)
or \(x^23=0\)
This is the required quadratic equation
 \(\frac{1}{2} (1+\sqrt{5}),\frac{1}{2} (1\sqrt{5}) \)
Solution 👉 Click Here
Solution :1d
Given roots roots are \(\frac{1}{2} (1+\sqrt{5}),\frac{1}{2} (1\sqrt{5}) \), thus
\( \text {sum of roots}=\left (\frac{1}{2} (1+\sqrt{5}) +\frac{1}{2} (1\sqrt{5}) \right)=1, \text {product of roots}=\left (\frac{1}{2} (1+\sqrt{5}).\frac{1}{2} (1\sqrt{5}) \right)=1\)
According to the question
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or \( x^2 \left ( 1 \right) x + \left ( 1 \right )=0\)
or \(x^2+x1=0\)
This is the required quadratic equation
 3+5i,35i
Solution 👉 Click Here
Solution :1e
Given roots roots are 3+5i,35i, thus
\( \text {sum of roots}=(3+5i) +(3+5i )=6, \text {product of roots}=(3+5i) (3+5i )=34\)
According to the question
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or \( x^2 \left ( 6 \right) x + \left ( 34 \right )=0\)
or \(x^2+6x+34=0\)
This is the required quadratic equation
 a+ib,aib
Solution 👉 Click Here
Solution :1f
Given roots roots are a+ib,aib, thus
\( \text {sum of roots}=(a+ib) +(aib )=2a, \text {product of roots}=(a+ib) +(aib)=a^2+b^2\)
According to the question
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or \( x^2 \left ( 2a \right) x + \left ( a^2+b^2 \right )=0\)
or \(x^2+2ax+a^2+b^2=0\)
This is the required quadratic equation

 Find a quadratic equation whose roots are twice the roots of \(4x^2+8x5=0\)
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Solution :2a
Given quadratic equation is
\(4x^2+8x5=0\)
Let \( \alpha,\beta\) are the roots of \(4x^2+8x5=0\), then
\( \alpha+ \beta=\frac{8}{4}=2, \alpha \beta=\frac{5}{4}\)
According to the question, \( 2\alpha,2\beta\) are the roots of new equation
So, for the new quadratic equation
\( \text {sum of roots}=(2\alpha)+ (2\beta)=2 (\alpha+ \beta)=4\)
\( \text {product of roots}=(2\alpha) (2\beta)=4 (\alpha \beta)=5\)
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or \( x^2 \left ( 4 \right) x + \left ( 5 \right )=0\)
or \(x^2+4x5=0\)
This is the required quadratic equation
 Find a quadratic equation whose roots are reciprocals of the roots of \(3x^25x2=0\)
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Solution :2(b)
Given quadratic equation is
\(3x^25x2=0\)
Let \( \alpha,\beta\) are the roots of \(3x^25x2=0\), then
\( \alpha+ \beta=\frac{5}{3}, \alpha \beta=\frac{2}{3}\)
According to the question, \( \frac{1}{\alpha},\frac{1}{\beta}\) are the roots of new equation
So, for the new quadratic equation
\( \text {sum of roots}=\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha + \beta}{\alpha \beta}=\frac{5}{2}\)
\( \text {product of roots}=\frac{1}{\alpha}\frac{1}{\beta}=\frac{1}{\alpha \beta}=\frac{3}{2}\)
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or \( x^2 \left (  \frac{5}{2} \right) x + \left ( \frac{3}{2} \right )=0\)
or \(2x^2+5x3=0\)
This is the required quadratic equation
 Find a quadratic equation whose roots are greater by h than the roots of \(x^2px+q=0\)
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Solution :2(c)
Given quadratic equation is
\(x^2px+q=0\)
Let \( \alpha,\beta\) are the roots of \(x^2px+q=0\), then
\( \alpha+ \beta=p, \alpha \beta=q\)
According to the question, \( \alpha +h,\beta +h\) are the roots of new equation
So, for the new quadratic equation
\( \text {sum of roots}=(\alpha+h)+(\beta+h)= 2h+ (\alpha+\beta)=2h+p\)
\( \text {product of roots}=(\alpha+h)(\beta+h)= \alpha \beta+h(\alpha+\beta)+h^2=q+ph+h^2\)
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or \( x^2 \left ( 2h+p \right) x + \left ( q+ph+h^2 \right )=0\)
This is the required quadratic equation
 Find a quadratic equation whose roots are the squares of the roots of \(3x^25x2=0\)
Solution 👉 Click Here
Solution :2(d)
Given quadratic equation is
\(3x^25x2=0\)
Let \( \alpha,\beta\) are the roots of \(3x^25x2=0\), then
\( \alpha+ \beta=\frac{5}{3}, \alpha \beta=\frac{2}{3}\)
According to the question, \( \alpha ^2,\beta ^2\) are the roots of new equation
So, for the new quadratic equation
\( \text {sum of roots}=(\alpha ^2)+(\beta ^2)= (\alpha+ \beta)^22 \alpha \beta=\left ( \frac{5}{3} \right )^2 2 \frac{2}{3}=\frac{37}{9}\)
\( \text {product of roots}=(\alpha ^2)(\beta ^2)= (\alpha \beta)^2=\frac{4}{9}\)
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or \( x^2 \left ( \frac{37}{9} \right) x + \left ( \frac{4}{9} \right )=0\)
or \( 9x^2 37 x + 4=0\)
This is the required quadratic equation
 Find a quadratic equation with rational coefficients one of whose roots is
 4+3i
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Solution :3a
Given that
one root is 4+3i
We know that, immaginary roots always occur in conjugates, therefore
other root is 43i
 \(\frac{1}{5+3i}\)
Solution 👉 Click Here
Solution :3b
Given that
one root is \(\frac{1}{5+3i}\)
or one root is \(\frac{1}{5+3i} \frac{53i}{53i}=\frac{53i}{34}\)
We know that, immaginary roots always occur in conjugates, therefore
other root is \( \frac{5+3i}{34}\)
 \(2+\sqrt{3}\)
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Solution :3c
Given that
one root is \(2+\sqrt{3}\)
We know that, immaginary roots always occur in conjugates, therefore
other root is \(2\sqrt{3}\)
 Find the value of k so that the equation
 \(2x^2+kx15=0\) has one root 3
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Solution :4a
Given quadratic equation is
\(2x^2+kx15=0\)
Comparing with \(ax^2+bx+c=0\), we get \(a=2,b=k,c=15\)
Also
one root is 3, and suppose next root is \(\alpha\)
Now
\(3+ \alpha=\frac{b}{a},3 \alpha =\frac{c}{a}\)
or \(3+ \alpha=\frac{k}{2},3 \alpha =\frac{15}{2}\)
or \(3+ \alpha=\frac{k}{2},\alpha =\frac{5}{2}\)
or \(3\frac{5}{2}=\frac{k}{2}\)
or \(\frac{1}{2}=\frac{k}{2}\)
or \(k=1\)
This completes the solution.
 \(3x^2+kx2=0\) has roots whose sum is equal to 6
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Solution :4b
Given quadratic equation is
\(3x^2+kx2=0\)
Comparing with \(ax^2+bx+c=0\), we get \(a=3,b=k,c=2\)
Also
sum of roots is 6
Now
\(\alpha+\beta=6\)
or \(\frac{k}{3}=6\)
or \(k=18\)
This completes the solution.
 \(2x^2+(4k)x17=0\) has roots equal but opposite in sign
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Solution :4c
Given quadratic equation is
\(2x^2+(4k)x17=0\)
Comparing with \(ax^2+bx+c=0\), we get \(a=2,b=4k,c=17\)
Also
roots are equal but opposite in sign
Now
\(\alpha=\beta\)
or \(\alpha +\beta =0\)
or \(\frac{b}{a}=0\)
or \(b=0\)
or \(4k=0\)
or \(k=4\)
This completes the solution.
 \(3x^2+(5+k)x+8=0\) has roots numerically equal but opposite in sign
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Solution :4d
Given quadratic equation is
\(3x^2+(5+k)x+8=0\)
Comparing with \(ax^2+bx+c=0\), we get \(a=3,b=5+k,c=8\)
Also
roots are equal but opposite in sign
Now
\(b=0\)
or \(5+k=0\)
or \(k=5\)
This completes the solution.
 \(3x^2+7x+6k=0\) has one root equal to zero
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Solution :4e
Given quadratic equation is
\(3x^2+7x+6k=0\)
Comparing with \(ax^2+bx+c=0\), we get \(a=3,b=7,c=6k\)
Also
one root is zero
Now
\(\alpha \beta =0\)
or \(\frac{c}{a}=0\)
or \(c=0\)
or \(6k=0\)
or \(k=6\)
This completes the solution.
 \(4x^217x+k=0\) has the reciprocal roots
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Solution :4f
Given quadratic equation is
\(4x^217x+k=0\)
Comparing with \(ax^2+bx+c=0\), we get \(a=4,b=17,c=k\)
Also
roots are in reciprocal relation
Now
\(\alpha = \frac{1}{\beta }\)
or \(\alpha \beta =1\)
or \(\frac{c}{a}=1\)
or \(\frac{k}{4}=1\)
or \(k=4\)
This completes the solution.
 \(4x^2+kx+5=0\) has roots whose difference is \(\frac{1}{4}\)
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Solution :4g
Given quadratic equation is
\(4x^2+kx+5=0\)
Comparing with \(ax^2+bx+c=0\), we get \(a=4,b=k,c=5\)
Also
difference of roots is 1/4
or \(\alpha  \beta =\frac{1}{4}\)
We know that
\(\alpha + \beta =\frac{k}{4},\alpha \beta =\frac{5}{4} \)
Thus,
\( (\alpha + \beta )^2= (\alpha \beta )^2+ 4 \alpha \beta \)
or \( \frac{k^2}{16}= \frac{1}{16}+5\)
or \(k= \pm 9\)
This completes the solution.
 Show that 1 is a root of the equation \((a+b2c)x^2+(2abc)x+(c+a2b)=0\). Find the other root.
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Solution :5
Given quadratic equation is
\((a+b2c)x^2+(2abc)x+(c+a2b)=0\)
 If (1) is aa root, then it musst satisfy the equation
\((a+b2c)x^2+(2abc)x+(c+a2b)=0\)
or \((a+b2c)(1)^2+(2abc)(1)+(c+a2b)=0\)
or \((a+b2c)(2abc)+(c+a2b)=0\)
or \(0=0\)
Therefore, we conclude that (1) is a root.
 Given that (1) is a root. Let us suppose that other root is \( \alpha\), then
\( \alpha (1)= \frac{c}{a}\)
or \( \alpha (1)= \frac{c+a2b}{a+b2c}\)
or \( \alpha =  \frac{c+a2b}{a+b2c}\)
this is the other root.
This completes the solution.
 Find the value of m for which the equation \((m+1)x^2+2(m+3)x+(2m+3)=0\) will have (a) reciprocal roots (b) one root zero.
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Solution :6

Given quadratic equation is
\((m+1)x^2+2(m+3)x+(2m+3)=0\)
Comparing with \(ax^2+bx+c=0\), we get \(a=m+1,b=2(m+3),c=2m+3\)
Also
roots are in reciprocal relation
Now
\(\alpha = \frac{1}{\beta }\)
or \(\alpha \beta =1\)
or \(\frac{c}{a}=1\)
or \(\frac{2m+3}{m+1}=1\)
or \(m=2\)
This completes the solution.
 Given quadratic equation is
\((m+1)x^2+2(m+3)x+(2m+3)=0\)
Comparing with \(ax^2+bx+c=0\), we get \(a=m+1,b=2(m+3),c=2m+3\)
Also
one root is zero
Now
\(\alpha \beta =0\)
or \(\frac{c}{a}=0\)
or \(\frac{2m+3}{m+1}=0\)
or \(m=3/2\)
This completes the solution.
 If the roots of the equation \(x^2+ax+c=0\) differ by 1, prove that \(a^2=4c+1\)
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Solution :7
Given quadratic equation is
\(x^2+ax+c=0\)
Let \(\alpha, \beta\) are the roots, then
\( \alpha+ \beta = \frac{b}{a} \) and \( \alpha \beta =\frac{c}{a} \)
or \( \alpha+ \beta =a \) and \( \alpha \beta =c \)
According to the question
\( \alpha \beta =1\)
or \( (\alpha \beta)^2=1 \)
or \( (\alpha+ \beta)^24 \alpha \beta =1 \)
or \( a^24 c =1 \)
or \( a^2=1 +4c\)
This completes the proof.
 If \( \alpha, \beta\) are the roots of the equation \(x^2x6=0\), find the equation whose roots are
 \(\alpha ^2 \beta ^{1}\)and \(\beta ^2 \alpha ^{1}\)
Solution 👉 Click Here
Solution :8a
Given quadratic equation is
\(x^2x6=0\)
Let \( \alpha,\beta\) are the roots of \(x^2x6=0\), then
\( \alpha+ \beta=\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
or
\( \alpha+ \beta=1, \alpha \beta=6\)
According to the question, the new roots are
\(\alpha ^2 \beta ^{1}\) and \(\beta ^2 \alpha ^{1}\)
So,
Sum of new roots 
=\(\alpha ^2 \beta ^{1}+ \beta ^2 \alpha ^{1}\) 

=\( \frac{\alpha ^2}{\beta}+ \frac{\beta ^2}{\alpha}\) 

=\( \frac{\alpha ^3+\beta ^3}{\alpha \beta}\)
=\( \frac{(\alpha +\beta) ^33\alpha \beta (\alpha +\beta) }{\alpha \beta}\)
=\( \frac{(\alpha +\beta) ^3} {\alpha \beta}  \frac{3\alpha \beta (\alpha +\beta) }{\alpha \beta}\)
=\( \frac{(\alpha +\beta) ^3} {\alpha \beta}  3 (\alpha +\beta) \)
=\( \frac{\left (1 \right ) ^3} {6}  3 \left (1 \right ) \)
=\( \frac{1}{6} 3 \)
=\( \frac{19}{6} \)

Again,
Product of new roots 
=\(\alpha ^2 \beta ^{1} . \beta ^2 \alpha ^{1}\) 

=\( \frac{\alpha ^2}{\beta}. \frac{\beta ^2}{\alpha}\) 

=\( \alpha \beta\)
=\( 6\)

Now
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or
\( x^2 \left ( \frac{19}{6} \right) x + \left (6 \right)=0\)
or
\( 6x^2+19x 36=0\)
This is the required quadratic equation
 \(\alpha + \frac{1}{\beta}\) and \(\beta + \frac{1}{\alpha}\)
Solution 👉 Click Here
Solution :8(b)
Given quadratic equation is
\(x^2x6=0\)
Let \( \alpha,\beta\) are the roots of \(x^2x6=0\), then
\( \alpha+ \beta=\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
or
\( \alpha+ \beta=1, \alpha \beta=6\)
According to the question, the new roots are
\(\alpha +\frac{1}{\beta}\) and \(\beta +\frac{1}{\alpha}\)
So,
Sum of new roots 
=\(\alpha +\frac{1}{\beta}+\beta +\frac{1}{\alpha}\) 

=\(\alpha +\beta +\frac{1}{\beta}+\frac{1}{\alpha} \) 

=\( (\alpha +\beta) +\frac{\alpha +\beta}{\alpha \beta} \)
=\( 1+\frac{1}{6}\)
=\(\frac{5}{6} \)

Again,
Product of new roots 
=\( \left ( \alpha +\frac{1}{\beta} \right ). \left ( \beta +\frac{1}{\alpha}\right ) \) 

=\(\frac{\alpha \beta+1}{\beta }\frac{\alpha \beta+1}{\alpha } \) 

=\(\frac{(\alpha \beta)^2+2(\alpha \beta) +1}{\alpha \beta } \)
=\(\frac{3612+1}{6}\)
=\(\frac{25}{6} \)

Now
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or
\( x^2 \left ( \frac{5}{6} \right) x + \left (\frac{25}{6} \right)=0\)
or
\( 6x^25x 25=0\)
This is the required quadratic equation
 If \( \alpha, \beta\) are the roots of the equation \(ax^2+bx+c=0\), find the equation whose roots are
 \(\alpha \beta ^{1}\)and \(\beta \alpha ^{1}\)
Solution 👉 Click Here
Solution :9a
Given quadratic equation is
\(ax^2+bx+c=0\)
Let \( \alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then
\( \alpha+ \beta=\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
According to the question, the new roots are
\(\alpha \beta ^{1}\) and \(\beta \alpha ^{1}\)
So,
Sum of new roots 
=\(\alpha \beta ^{1}+ \beta \alpha ^{1}\) 

=\( \frac{\alpha }{\beta}+ \frac{\beta }{\alpha}\) 

=\( \frac{\alpha ^2+\beta ^2}{\alpha \beta}\)
=\( \frac{(\alpha +\beta) ^22\alpha \beta }{\alpha \beta}\)
=\( \frac{(\alpha +\beta) ^2}{\alpha \beta} 2 \frac {\alpha \beta }{\alpha \beta}\)
=\( \frac{(\alpha +\beta) ^2}{\alpha \beta} 2 \)
=\( \frac{b^22ac}{ac} \)

Again,
Product of new roots 
=\(\alpha \beta ^{1} . \beta \alpha ^{1}\) 

=\( \frac{\alpha }{\beta}. \frac{\beta }{\alpha}\) 

=\( 1\)

Now
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or
\( x^2 \left ( \frac{b^22ac}{ac} \right) x + \left (1 \right)=0\)
or
\( acx^2 (b^22ac)x + ac=0\)
This is the required quadratic equation
 \(\alpha ^3 \) and \(\beta ^3\)
Solution 👉 Click Here
Solution :9b
Given quadratic equation is
\(ax^2+bx+c=0\)
Let \( \alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then
\( \alpha+ \beta=\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
According to the question, the new roots are
\(\alpha ^3 \) and \(\beta ^3\)
So,
Sum of new roots 
=\(\alpha ^3+ \beta ^3\) 

=\( (\alpha +\beta) ^33\alpha \beta (\alpha +\beta)\) 

=\( \left (\frac{b}{a} \right ) ^3 3 \left (\frac{c}{a} \right ) \left (\frac{b}{a} \right ) \)
=\( \frac{b^3}{a^3} + \frac{3bc}{a^2} \)
=\( \frac{3abcb^3}{a^3} \)

Again,
Product of new roots 
=\(\alpha ^3 .\beta ^3\) 

=\( \left ( \frac{c}{a} \right)^3\) 

=\( \frac{c^3}{a^3}\)

Now
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or
\( x^2 \left ( \frac{3abcb^3}{a^3} \right) x + \left ( \frac{c^3}{a^3} \right)=0\)
or
\( a^3x^2+ (b^33abc)x + c^3=0\)
This is the required quadratic equation
 \((\alpha\beta)^2 \) and \((\alpha+\beta)^2 \)
Solution 👉 Click Here
Solution :9c
Given quadratic equation is
\(ax^2+bx+c=0\)
Let \( \alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then
\( \alpha+ \beta=\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
According to the question, the new roots are
\((\alpha\beta)^2 \) and \((\alpha+\beta)^2 \)
So,
Sum of new roots 
=\((\alpha+\beta)^2 +(\alpha\beta)^2 \) 

=\((\alpha+\beta)^2 +(\alpha+\beta)^2 4 \alpha \beta \) 

=\(2 \left ( \frac{b}{a}\right )^2 4 \left ( \frac{c}{a}\right ) \)
=\(\frac{2b^24ac}{a^2} \)

Again,
Product of new roots 
=\((\alpha+\beta)^2 .(\alpha\beta)^2 \) 

=\((\alpha+\beta)^2 \left [(\alpha+\beta)^2 4 \alpha \beta \right ]\) 

=\((\alpha+\beta)^4 4 (\alpha \beta )(\alpha+\beta)^2 \)
=\(\frac{b^4}{a^4} 4 \frac{c}{a} \frac{b^2}{a^2} \)
=\(\frac{b^44a^2b^2c}{a^4} \)

Now
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or
\( x^2 \left ( \frac{2b^24ac}{a^2} \right) x + \left (\frac{b^44a^2b^2c}{a^4} \right)=0\)
or
\( a^4x^2 (2a^2b^24a^3c)x + (b^44a^2b^2c)=0\)
This is the required quadratic equation
 the reciprocal of the roots of given equation
Solution 👉 Click Here
Solution :9d
Given quadratic equation is
\(ax^2+bx+c=0\)
Let \( \alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then
\( \alpha+ \beta=\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
According to the question, the new roots are
\( \frac{1}{\alpha} \) and \(\frac{1}{\beta} \)
So,
Sum of new roots 
=\(\frac{1}{\alpha}+\frac{1}{\beta} \) 

=\(\frac{\alpha+\beta}{\alpha \beta} \) 

=\( \frac{b}{c} \)

Again,
Product of new roots 
=\(\frac{1}{\alpha} \frac{1}{\beta}\) 

=\(\frac{1}{\alpha \beta}\) 

=\(\frac{a}{c} \)

Now
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or
\( x^2 \left ( \frac{b}{c} \right) x + \left (\frac{a}{c} \right)=0\)
or
\( cx^2+bx +a=0\)
This is the required quadratic equation

 If the roots of the equation \(ax^2+bx+c=0\) be in the ratio of 3:4, prove that \(12b^2=49ac\)
Solution 👉 Click Here
Solution :10a
Given quadratic equation is
\(ax^2+bx+c=0\)
Assume that \( 3k,4k\) are the roots of \(ax^2+bx+c=0\), then
\( 3k+ 4k=\frac{b}{a}, 3k.4k=\frac{c}{a}\)
or \( 7k=\frac{b}{a}, 12k^2=\frac{c}{a}\)
or \( k=\frac{b}{7a}, 12k^2=\frac{c}{a}\)
or \( 12 \left ( \frac{b}{7a} \right )^2=\frac{c}{a}\)
or \( 12 \left ( \frac{b^2}{49a^2} \right )=\frac{c}{a}\)
or \( 12 b^2=49ac\)
This completes the proof.
 If one root of the equation \(ax^2+bx+c=0\) be four times the other root, show that \(4b^2=25ac\)
Solution 👉 Click Here
Solution :10b
Given quadratic equation is
\(ax^2+bx+c=0\)
Assume that \( k,4k\) are the roots of \(ax^2+bx+c=0\), then
\( k+ 4k=\frac{b}{a}, k.4k=\frac{c}{a}\)
or \( 5k=\frac{b}{a}, 4k^2=\frac{c}{a}\)
or \( k=\frac{b}{5a}, 4k^2=\frac{c}{a}\)
or \( 4 \left ( \frac{b}{5a} \right )^2=\frac{c}{a}\)
or \( 4 \left ( \frac{b^2}{25a^2} \right )=\frac{c}{a}\)
or \( 4 b^2=25ac\)
This completes the proof.
 For what values of m, the equation \(x^2mx+m+1=0\) may have its root in the ratio 2:3
Solution 👉 Click Here
Solution :10c
Given quadratic equation is
\(x^2mxx+m+1=0\)
Assume that \( 2k,3k\) are the roots of \(x^2mxx+m+1=0\), then
\( 2k+ 3k=m, 2k.3k=m+1\)
Now, using the roots, we get
\(6k^2=m+1\)
or\(6 \left (\frac{m}{5} \right )^2 =m+1\)
or\(\frac{6m^2}{25} =m+1\)
or\(6m^225m25=0\)
or\(m=5,\frac{5}{6}\)
This completes the proof

 If \( \alpha, \beta\) are the roots of the equation \(px^2+qx+q=0\), prove that \(\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{q}{p}}=0\)
Solution 👉 Click Here
Solution :11a
Method 1
Given quadratic equation is
\(px^2+qx+q=0\)
Given that \( \alpha, \beta\) are the roots of \(px^2+qx+q=0\), then
\( \alpha+ \beta=\frac{q}{p}, \alpha \beta=\frac{q}{p}\)
Now, using the roots, we get
\(\frac{\alpha+ \beta}{\sqrt{\alpha \beta}}=\frac{\frac{q}{p}}{\frac{q}{p}}\)
or\(\frac{\alpha}{\sqrt{\alpha \beta}}+\frac{ \beta}{\sqrt{\alpha \beta}}= \sqrt{\frac{q}{p}}\)
or\( \sqrt{ \frac{\alpha}{\beta}}+ \sqrt{ \frac{ \beta}{\alpha }}+ \sqrt{\frac{q}{p}}=0\)
This completes the proof
Method 2
Given quadratic equation is
\(px^2+qx+q=0\)
Given that \( \alpha, \beta\) are the roots of \(px^2+qx+q=0\), then
\( \alpha+ \beta=\frac{q}{p}, \alpha \beta=\frac{q}{p}\)
Now, taking LHS, we have
\( \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{q}{p}}\)
or\( \frac{\sqrt{\alpha}}{\sqrt{\beta}}+\frac{\sqrt{\beta}}{\sqrt{\alpha}}+\sqrt{\frac{q}{p}}\)
or\( \frac{\alpha+\beta}{\sqrt{\alpha \beta}}+\sqrt{\frac{q}{p}}\)
or\( \frac{\frac{q}{p}}{\sqrt{\frac{q}{p}}}+\sqrt{\frac{q}{p}}\)
or\( \sqrt{\frac{q}{p}}+\sqrt{\frac{q}{p}}\)
or0
Which is RHS, therefore, we have
\( \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{q}{p}}=0\)
 If roots of the equation \(lx^2+nx+n=0\) be in the ratio of p:q, prove that \(\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0\)
Solution 👉 Click Here
Solution :11b
Method 0
Given quadratic equation is
\(lx^2+nx+n=0\)
Given that \( \alpha, \beta\) are the roots of \(lx^2+nx+n=0\), then
\( \alpha+ \beta=\frac{n}{l}, \alpha \beta=\frac{n}{l}, \frac{\alpha}{\beta}=\frac{p}{q}\)
Now, using the roots, we get
\(\frac{\alpha+ \beta}{\sqrt{\alpha \beta}}=\frac{\frac{n}{l}}{ \sqrt{\frac{n}{l}}}\)
or\(\frac{\alpha}{\sqrt{\alpha \beta}}+\frac{ \beta}{\sqrt{\alpha \beta}}= \sqrt{\frac{n}{l}}\)
or\( \sqrt{ \frac{\alpha}{\beta}}+ \sqrt{ \frac{ \beta}{\alpha }}+ \sqrt{\frac{n}{l}}=0\)
or\( \sqrt{ \frac{p}{q}}+ \sqrt{ \frac{ q}{p}}+ \sqrt{\frac{n}{l}}=0\)
This completes the proof
Method 1
Given quadratic equation is
\(lx^2+nx+n=0\)
Let us assume that \( p \alpha,q \alpha \) are the roots of \(lx^2+nx+n=0\), then the roots are in the ration of p:q, now according to the question
\(p \alpha q \alpha =\frac{n}{l}\)
or \(pq \alpha ^2 =\frac{n}{pql}\)
Since \(p \alpha\) is the one root of the equation \(lx^2+nx+n=0\), we get
\(l (p \alpha)^2+n(p \alpha)+n=0\)
or \(lp^2 (\alpha ^2) +np(\alpha)+n=0\)
or \(lp^2 \frac{n}{pql} +np \sqrt{\frac{n}{pql}} +n=0\)
or \(\frac{p}{q} +\sqrt{\frac{np}{ql}} +1=0\)
Dividing both side by \(\sqrt{\frac{p}{q}} \), we get
\(\sqrt{\frac{p}{q}} +\sqrt{\frac{n}{l}} +\sqrt{\frac{q}{p}}=0\)
or \(\sqrt{\frac{p}{q}} +\sqrt{\frac{q}{p}} +\sqrt{\frac{n}{l}} =0\)
Method 2
Given quadratic equation is
\(lx^2+nx+n=0\)
Let \( \alpha, \beta\) are the roots of \(lx^2+nx+n=0\), then
\( \alpha+ \beta=\frac{n}{l}, \alpha \beta=\frac{n}{l},\frac{\alpha}{ \beta}=\frac{p}{q}\)
Now, taking LHS, we have
\( \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}\)
or\( \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{n}{l}}\)
or\( \frac{\sqrt{\alpha}}{\sqrt{\beta}}+\frac{\sqrt{\beta}}{\sqrt{\alpha}}+\sqrt{\frac{n}{l}}\)
or\( \frac{\alpha+\beta}{\sqrt{\alpha \beta}}+\sqrt{\frac{n}{l}}\)
or\( \frac{\frac{n}{l}}{\sqrt{\frac{n}{l}}}+\sqrt{\frac{n}{l}}\)
or\( \sqrt{\frac{n}{l}}+\sqrt{\frac{n}{l}}\)
or0
Which is RHS, therefore, we have
\( \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0\)
 If one root of the equation \(ax^2+bx+c=0\) be square of the other root, prove that \(b^3+a^2c+ac^2=3abc\)
Solution 👉 Click Here
Solution :12
Given quadratic equation is
\(ax^2+bx+c=0\)
Let us assume that \( \alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then
According to question
\(\alpha=\beta ^2, \alpha+\beta =\frac{b}{a}, \alpha\beta =\frac{c}{a}\)
or \(\beta ^2+\beta =\frac{b}{a}, \beta ^2\beta =\frac{c}{a}\)
or \(\beta (1+ \beta) =\frac{b}{a}, \beta ^3 =\frac{c}{a}\)
Taking the cube of first part, we get
\(\beta ^3 (1+ \beta)^3 =\frac{b^3}{a^3}\)
or \(\frac{c}{a} (1+ 3 \beta+3 \beta ^2+\beta ^3) =\frac{b^3}{a^3}\)
or \(\frac{c}{a} (1+ 3 \beta (1+\beta) +\beta ^3) =\frac{b^3}{a^3}\)
or \(\frac{c}{a} \left ( 1+ 3 \left ( \frac{b}{a} \right ) +\frac{c}{a} \right ) =\frac{b^3}{a^3}\)
or \(\frac{c}{a} \left ( \frac{a3b+c}{a} \right ) =\frac{b^3}{a^3}\)
or \(\frac{ac3bc+c^2}{a^2} =\frac{b^3}{a^3}\)
or \(a^2c3abc+ac^2 =b^3\)
or \(b^3+a^2c+ac^2 =3abc\)
Grade 11 Mathematics: Quadratic Equation [BCB Exercise 6.3]
 Show that each pair of following equations has a common root
 \( x^28x+15=0\) and \(2x^2x15 =0\)
Solution 👉 Click Here
Solution :1(a)
Given quadratic equations are
\( x^28x+15=0\) and \(2x^2x15 =0\)
Comparing with \( a_1x^2+b_1x+c_1=0\) and \( a_2x^2+b_2x+c_2=0\), we get
\(a_1=1,b_1=8,c_1=15,a_2=2,b_2=1,c_2=15\)
We know that, condition for two quadratic equations having a common root is
\( (a_2c_1a_1c_2)^2 = (a_1b_2a_2b_1) (b_1c_2b_2c_1) \)
or \( [2.(15)1(15)]^2 = [1(1)2(8)] [(8)(15)(1)(15)]\)
or \( [30+15]^2 = [1+16] [120+15]\)
or \(2035 =2035\)
Thus, \( x^28x+15=0\) and \(2x^2x15 =0\) has a common root.
 \( 3x^28x+4=0\) and \(4x^27x2 =0\)
Solution 👉 Click Here
Solution :1(b)
Given quadratic equations are
\( 3x^28x+4=0\) and \(4x^27x2 =0\)
Comparing with \( a_1x^2+b_1x+c_1=0\) and \( a_2x^2+b_2x+c_2=0\), we get
\( a_1=3,b_1=8,c_1=4,a_2=4,b_2=7,c_2=2\)
We know that, condition for two quadratic equations having a common root is
\( (a_2c_1a_1c_2)^2 = (a_1b_2a_2b_1) (b_1c_2b_2c_1) \)
or \( [(4)(4)(3)(2)]^2 = [(3)(7)(4)(8)] [(8)(2)(7)(4)] \)
or \( [16+6]^2 = [21+32] [16+28]\)
or \(484 =484\)
Thus, \( 3x^28x+4=0\) and \(4x^27x2 =0\) has a common root.
 Find the value of p so that each pair of the equations may have one root common
 \( 4x^2+px12=0\) and \(4x^2+3px4 =0\)
Solution 👉 Click Here
Solution :2(a)
Given quadratic equations are
\( 4x^2+px12=0\) and \(4x^2+3px4 =0\)
Comparing with \( a_1x^2+b_1x+c_1=0\) and \( a_2x^2+b_2x+c_2=0\), we get
\( a_1=4,b_1=p,c_1=12,a_2=4,b_2=3p,c_2=4\)
We know that, condition for two quadratic equations having a common root is
\( (c_1a_2c_2a_1)^2 = (a_1b_2a_2b_1) (b_1c_2b_2c_1) \)
or \( [(12)(4)(4)(4)]^2 = [(4)(3p)(4)(p)] [(p)(4)(3p)(12)]\)
or \( [48+16]^2 = [12p4p] [4p+36p]\)
or \( [32]^2 = [8p] [32p]\)
or \( p^2=4\)
or \(p=\pm 2\)
Thus,fo \(p=\pm 2\), given pair of the equations may have one root common
 \( 2x^2+px1=0\) and \(3x^22x5 =0\)
Solution 👉 Click Here
Solution :2(b)
Given quadratic equations are
\(2x^2+px1=0\) and \(3x^22x5=0\)
Comparing with \( a_1x^2+b_1x+c_1=0\) and \( a_2x^2+b_2x+c_2=0\), we get
\( a_1=2,b_1=p,c_1=1,a_2=3,b_2=2,c_2=5\)
We know that, condition for two quadratic equations having a common root is
\( (c_1a_2c_2a_1)^2 = (a_1b_2a_2b_1) (b_1c_2b_2c_1) \)
or \( [(1)(3)(5)(2)]^2 = [(2)(2)(3)(p)] [(p)(5)(2)(1)]\)
or \( [3+10]^2 = [43p] [5p2]\)
or \( [7]^2 = [4+3p] [2+5p]\)
or \(49 = 15p^2+26p+8\)
or \( 15p^2+26p41=0\)
or \( 15p^2+(4115)p41=0\)
or \( 15p^2+41p15p41=0\)
or \( p(15p+41)1(15p+41)=0\)
or \( p=1, \frac{41}{15}\)
Thus,fo \( p=1, \frac{41}{15}\), given pair of the equations may have one root common
 If the quadratic equations \(x^2+px+q=0\) and \(x^2+p'x+q'=0 \) have a common root show that the root must be either \(\frac{pq'p'q}{qq'} \) or \(\frac{qq'}{p'p} \)
Solution 👉 Click Here
Solution :3
Given quadratic equations are
\(x^2+px+q=0\) and \(x^2+p'x+q'=0\)
Let \( \alpha \) be the common root of the equation \(x^2+px+q=0\) and \(x^2+p'x+q'=0\)then
\( \alpha ^2+p \alpha +q=0\) (1)
\(\alpha ^2+p' \alpha +q'=0\)(2)
Substracting (2) from (1), we get
\((pp') \alpha +(qq')=0\)
or \( \alpha= \frac{q'q}{pp'} \)
or \( \alpha= \frac{qq'}{p'p} \)
Multiplying (1) by \(q'\), (2) by \(q\) and substracting, we get
\((q'q) \alpha +(pq'p'q)=0\)
or \( \alpha= \frac{pq'p'q}{qq'} \)
Thus,\(x^2+px+q=0\) and \(x^2+p'x+q'=0\) have one common root \(\alpha\) if \( \alpha= \frac{qq'}{p'p} \) or \( \alpha= \frac{pq'p'q}{qq'} \)
 If the quadratic equations \(x^2+px+q=0\) and \(x^2+qx+p=0 \) have common roots show that it must be either \( p=q\) or \( p+q+1=0\)
Solution 👉 Click Here
Solution :4
Given quadratic equations are
\(x^2+px+q=0\) and \(x^2+qx+p=0\)
Comparing with \( a_1x^2+b_1x+c_1=0\) and \( a_2x^2+b_2x+c_2=0\), we get
\( a_1=1,b_1=p,c_1=q,a_2=1,b_2=q,c_2=p\)
We know that, condition for two quadratic equations having a common root is
\( (c_1a_2c_2a_1)^2 = (a_1b_2a_2b_1) (b_1c_2b_2c_1) \)
or \( [(q)(1)(p)(1)]^2 = [(1)(q)(1)(p)] [(p)(p)(q)(q)]\)
or \( [qp]^2 = [qp] [p^2q^2]\)
or \( (qp) [(qp) (p^2q^2)]=0\)
or \( qp =0\) or \( p+q1 =0\)
Thus,\(x^2+px+q=0\) and \(x^2+qx+p=0\) have one root common if \( p=q\) or \( p+q1 =0\)
 If the quadratic equations \(ax^2+bx+c=0\) and \(bx^2+cx+a=0 \) have common roots show that it must be either \( a=b=c\) or \( a+b+c=0\)
Solution 👉 Click Here
Solution :5
Given quadratic equations are
\(ax^2+bx+c=0\) and \(bx^2+cx+a=0\)
Comparing with \( a_1x^2+b_1x+c_1=0\) and \( a_2x^2+b_2x+c_2=0\), we get
\( a_1=a,b_1=b,c_1=c,a_2=b,b_2=c,c_2=a\)
We know that, condition for two quadratic equations having a common root is
\( (c_1a_2c_2a_1)^2 = (a_1b_2a_2b_1) (b_1c_2b_2c_1) \)
or \( [(c)(b)(a)(a)]^2 = [(a)(c)(b)(b)] [(b)(a)(c)(c)]\)
or \( [bca^2]^2 = [cab^2] [abc^2]\)
or \( b^2c^22a^2bc+a^4=a^2bcac^3ab^3+b^2c^2\)
or \( a^4=3a^2bcac^3ab^3\)
or \( a^3=3abcc^3b^3\)
or \( a^3+b^3+c^33abc=0\)
or \( a^3+b^3+c^33abc=0\)
or \( (a+b+c)(a^2+b^2+c^2abbcca)=0\)
or Either \( (a+b+c)=0\) or \((a^2+b^2+c^2abbcca)=0\)
or Either \( a+b+c=0\) or \(2a^2+2b^2+2c^22ab2bc2ca=0\)
or Either \( a+b+c=0\) or \((ab)^2+(bc)^2+(ca)^2=0\)
or Either \( a+b+c=0\) or \(ab=0, bc=0,ca\)
or Either \( a+b+c=0\) or \(a=b,b=c,c=a\)
or Either \( a+b+c=0\) or \(a=b=c\)
Thus,\(ax^2+bx+c=0\) and \(bx^2+cx+a=0\) have one root common if \( a+b+c=0\) or \(a=b=c\)
 Prove that if the equations \(x^2+bx+ca=0\) and \(x^2+cx+ab=0 \) have a common root, their other root will satisfy \( x^2+ax+bc=0\)
Solution 👉 Click Here
Solution :6
Given quadratic equations are
\(x^2+bx+ca=0\) and \(x^2+cx+ab=0\)
Let \(k\) be the common solution of the equations \(x^2+bx+ca=0\) and \(x^2+cx+ab=0\) then
\(k^2+bk+ca=0\) and \(k^2+ck+ab=0\)
Solving the both equations for k, we get
\(\frac{k^2}{ \begin{vmatrix} b & ca \\ c & ab \end{vmatrix} }=\frac{k}{\begin{vmatrix} ca & 1 \\ ab & 1 \end{vmatrix}}=\frac{1}{\begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix}} \)
Comparing first two, we get
\(k=\frac{ \begin{vmatrix} b & ca \\ c & ab \end{vmatrix} } {\begin{vmatrix} ca & 1 \\ ab & 1 \end{vmatrix} } \)
or\(k=\frac{ab^2ac^2}{acab} \)
or\(k=(b+c) \)
Again, comparing last two, we get
\(k= \frac{\begin{vmatrix} ca & 1 \\ ab & 1 \end{vmatrix}}{\begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix}} \)
or\(k=\frac{acab}{cb} \)
or\(k=a \)
Using both values of k, we get
\(a=(b+c) \)
or\(a+b+c=0 \)
Now, suppose \(\alpha \) be the other root of \(x^2+bx+ca=0\), then
\( k+\alpha=b\)
or\( a+\alpha=b\)
or\( \alpha=ab\)
Next, suppose \(\beta \) be the other root of \(x^2+cx+ab=0\), then
\( k+\beta=c\)
or\( a+\beta=c\)
or\( \beta=ac\)
Now, new quadratic equation containing other roots \(\alpha, \beta\)is given by
\(x^2(\alpha+ \beta)x+ (\alpha \beta)=0\)
or\(x^2(abac)x+ (ab)(ac)=0\)
or\(x^2(2abc)x+ (a^2abac+bc)=0\)
or\(x^2(2a+a)x+ (0+bc)=0\)
or\(x^2+ax+ bc=0\)
Extra Question
If \( \alpha, \beta\) are the roots of the equation \(ax^2+bx+c=0\), find the equation whose roots are
\(\alpha ^2 \beta ^{1}\)and \(\beta ^2 \alpha ^{1}\)
Solution
Given quadratic equation is
\(ax^2+bx+c=0\)
Let \( \alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then
\( \alpha+ \beta=\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
According to the question, the new roots are
\(\alpha ^2 \beta ^{1}\) and \(\beta ^2 \alpha ^{1}\)
So,
Sum of new roots 
=\(\alpha ^2 \beta ^{1}+ \beta ^2 \alpha ^{1}\) 

=\( \frac{\alpha ^2}{\beta}+ \frac{\beta ^2}{\alpha}\) 

=\( \frac{\alpha ^3+\beta ^3}{\alpha \beta}\)
=\( \frac{(\alpha +\beta) ^33\alpha \beta (\alpha +\beta) }{\alpha \beta}\)
=\( \frac{(\alpha +\beta) ^3} {\alpha \beta}  \frac{3\alpha \beta (\alpha +\beta) }{\alpha \beta}\)
=\( \frac{(\alpha +\beta) ^3} {\alpha \beta}  3 (\alpha +\beta) \)
=\( \frac{\left (\frac{b}{a} \right ) ^3} {\frac{c}{a}}  3 \left (\frac{b}{a} \right ) \)
=\( \frac{b^3}{ca^2} + \frac{3b}{a} \)
=\( \frac{3abcb^3}{a^2c} \)

Again,
Product of new roots 
=\(\alpha ^2 \beta ^{1} . \beta ^2 \alpha ^{1}\) 

=\( \frac{\alpha ^2}{\beta}. \frac{\beta ^2}{\alpha}\) 

=\( \alpha \beta\)
=\( \frac{c}{a}\)

Now
The new quadratic equation is
\( x^2  (\text {sum of roots}) x + (\text {product of roots}) =0\)
or \( x^2 \left ( \frac{3abcb^3}{a^2c} \right) x + \left ( \frac{c}{a} \right)=0\)
or \( a^2cx^2+ (b^33abc)x + ac^2=0\)
This is the required quadratic equation
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