Quadratic Equation

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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.
  1. The standard form of Quadratic Equation is
    \( ax^2 + bx + c = 0\) where \(a \ne 0\)
  2. Because the quadratic equation involves only one unknown, it is called "univariate".
  3. The numbers a, b, and c are the coefficients, they are also respectively called quadratic coefficient, linear coefficient, and the constant or free term.
  4. The quadratic equation contains only powers of x that are non-negative integers, and therefore it is a polynomial equation. In particular, it is a second-degree polynomial equation, since the greatest power is two.
  5. 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.
  6. 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.
  7. 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.
  8. A quadratic equation always has two roots, if complex roots are included and a double root is counted for two.
  9. A quadratic polynomial has a parabola as its graph

Three forms of Quaadratic Equation

We can obtain three different forms of a quadratic function

  1. 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 \(b^2-4ac\) is positive, zero, or negative, respectively
  2. The Vertex form representation of Quadratic Equation is : \( f (x) = a(x-p)^2 + q\). 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 \(x = p\).
    1. If a > 0, f(x) is decreasing on (−∞, p] and therefore increasing on [p,∞).
    2. If a < 0, then f(x) is increasing on (−∞, p] and therefore decreasing on [p,∞).
    3. If a > 0, then f(x) achieves its minimum at p.
    4. If a < 0, then f(x) achieves its maximum at p.
  3. 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 well-rounded 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:
  1. Graphical Method
  2. Factoring Method
  3. 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
    \( x= \frac{-b \pm \sqrt {b^2-4ac}}{2a} \)
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 = 2x2 − x − 3 intersects the x-axis at (−1, 0) and (1.5, 0), and a simple computation confirms that −1 and 1.5 are roots of 2x2 − x − 3 = 0

Formula Method

Prove that two roots of quadratic equations \(ax^2 + bx + c = 0\) are \(\frac{-b - \sqrt{b^2-4ac}}{2a}\) and \(\frac{-b + \sqrt{b^2-4ac}}{2a}\)
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 \(\left( x+\frac{b}{2a} \right )^2=\frac{b^2-4ac}{4a^2}\)
Taking square roots we get
\(x+\frac{b}{2a}=\frac{\pm \sqrt{b^2-4ac}}{2a}\)
or \(x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)
So, the roots of quadratic equations \(ax^2 + bx + c = 0\) are \(\frac{-b - \sqrt{b^2-4ac}}{2a}\) and \(\frac{-b + \sqrt{b^2-4ac}}{2a}\)

To solve a Quadratic equation using the formula, we use the following steps:
  1. Put the quadratic equation into standard form : \(ax^2 + bx + c = 0\)
  2. Write out the value for a, b, and c
  3. Substitute value in the formula, solve and get the roots
  4. Check each root in the original equation
Nature of Roots

In the quadratic equation \(ax^2+bx+c=0\), the expression \(b^2-4ac\) is called the discriminant, and is often represented using an upper case D or an upper case Greek delta \( \Delta \) given as
\( \Delta=b^2-4ac \)
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:

  1. When the discriminant \( \Delta\) is positive, the curve of quadratic equation (parabola) intersects the x-axis 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,
    1. if the discriminant is a square number, then the roots are rational
    2. 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
      \( \alpha= \frac {-b+\sqrt{ b^2-4ac }}{2a}\) and \( \beta= \frac {-b-\sqrt{ b^2-4ac }}{2a}\)
      or \( \alpha= \frac {-b}{2a}+\frac{\sqrt{ b^2-4ac }}{2a}\) and \( \beta= \frac {-b}{2a}-\frac{\sqrt{ b^2-4ac }}{2a}\)
      or \( \alpha= p+\sqrt{q}\) and \( \beta= p-\sqrt{q}\)
  2. When the discriminant \( \Delta\) is zero, the vertex of the curve of quadratic equation (parabola) touches the x-axis at a single point. Then there is exactly one real root
    \( \frac{-b}{2a}\)
    sometimes called a repeated or double root.
  3. When the discriminant \( \Delta\) is negative, the curve of quadratic equation (parabola) does not intersect the x-axis at all. Then there are no real roots. Rather, there are two distinct (non-real) complex roots
    This kinds of complex roots always occur in conjugate pairs, which are of the form
    \( \alpha= \frac {-b+\sqrt{ b^2-4ac }}{2a}\) and \( \beta= \frac {-b-\sqrt{ b^2-4ac }}{2a}\)
    or \( \alpha= \frac {-b}{2a}+\frac{\sqrt{ b^2-4ac }}{2a}\) and \( \beta= \frac {-b}{2a}-\frac{\sqrt{ b^2-4ac }}{2a}\)
    or \( \alpha= p+i q\) and \( \beta= p-iq\)
    In these expressions i is the imaginary unit.
  4. Thus the roots are distinct if and only if the discriminant is non-zero, and the roots are real if and only if the discriminant is non-negative.

Based on the discriminant \( b^2-4ac\) of the quadratic equation \( ax^2+bx+c=0\) , the corresponding solution will be

If the discriminant is then there (roots)
positive \( b^2-4ac >0\) and a perfect square are two rational roots
positive \( b^2-4ac >0\) and NOT a perfect square are two irrational roots
zero: \( b^2-4ac=0\) is one rational roots
negative: \( b^2-4ac < 0\) are two complex roots

Graph of Quadratic Equation

The graph of a quadratic equation is a U-shaped 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.

  1. 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 \(f(x)=a(x-h)^2+k\), where (h,k) is the vertex.
    1. the h-coordinate of the vertex is given by \(h=-\frac{b}{2a}\)
    2. the k-coordinate of the vertex is given by \(k=f(h)= f\left (-\frac{b}{2a} \right )\)
    3. 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 \(b^2-4ac\), there are six possibilities in the graph.
a>0
Δ is positive
\(x^2-5x+4=0\)		 Δ is zero
\(x^2-6x+9=0\)		 Δ is negative
\(x^2-4x+5=0\)
a<0
Δ is positive
\(-x^2-5x-4=0\)		 Δ is zero
\(-x^2-6x-9=0\)		 Δ is negative
\(-x^2-4x-5=0\)
Note: the example of the above graph are given below.
graph of a quadratic function \(b^2-4ac > 0\) \(b^2-4ac=0\) \(b^2-4ac < 0 \)
\(a> 0\) 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 x-axis 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 x-axis.
  1. The quadratic function f(x) will be positive, i.e., f(x) > 0, for the values of x lying in the interval (-∞, α) ∪ (β, ∞)
  2. The quadratic function f(x) will be equal to zero, i.e., f(x) = 0, if x = α or β
  3. 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 x-axis 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 x-axis. The quadratic equation will have imaginary roots, and the curve will always lie above the x-axis. 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 x-axis 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 x-axis.
  1. The quadratic function f(x) will be negative, i.e., f(x) > 0, for the values of x lying in the interval (-∞, α) ∪ (β, ∞)
  2. The quadratic function f(x) will be equal to zero, i.e., f(x) = 0, if x = α or β
  3. 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 x-axis 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 x-axis. The quadratic equation will have imaginary roots, and the curve will always lie below the x-axis. 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
\( \alpha=\frac{-b - \sqrt{b^2-4ac}}{2a}\) and \(\beta =\frac{-b + \sqrt{b^2-4ac}}{2a}\)
By addition, we get that
\( \alpha+\beta=\frac{-b - \sqrt{b^2-4ac}}{2a}+\frac{-b + \sqrt{b^2-4ac}}{2a}\)
or\( \alpha+\beta=\frac{-2b}{2a}\)
or\( \alpha+\beta=-\frac{b}{a}\)
Again, By multiplication, we get that
\( \alpha. \beta=\frac{-b - \sqrt{b^2-4ac}}{2a}. \frac{-b + \sqrt{b^2-4ac}}{2a}\)
or\( \alpha\beta=\frac{(-b)^2-(b^2-4ac)^2}{4a^2}\)
or\( \alpha \beta=\frac{c}{a}\)

Some special cases
  1. 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: \(ax^2 + c = 0\).
  2. 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: \(ax^2 + bx = 0\).
  3. 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: \(ax^2 = 0\).

Formation of Quadratic Equation

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

  1. an expression in α and β which remains same when α and β are interchanged
  2. an expression in α and β which remains same when multiplied by -1, when α and β are interchanged

      3. 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
      1. \( \alpha + \beta\)
      2. \( \alpha \beta\)
      3. \( \alpha - \beta\)
        because \( \alpha - \beta =\sqrt{(\alpha + \beta)^2-4\alpha \beta }\)
        means \( \alpha - \beta\) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
      4. \( \alpha ^2 + \beta ^2\)
        because \( \alpha ^2 + \beta ^2 =(\alpha + \beta)^2-2\alpha \beta \)
        means \( \alpha ^2 + \beta ^2\) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
      5. \( \alpha ^2 - \beta ^2\)
        because \( \alpha ^2 - \beta ^2 = (\alpha + \beta) \sqrt{(\alpha + \beta)^2-4\alpha \beta }\)
        means \( \alpha ^2 - \beta ^2 \) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
      6. \( (\alpha - \beta )^2\)
        because \( (\alpha - \beta )^2 =(\alpha + \beta)^2-4\alpha \beta \)
        means \( (\alpha - \beta )^2 \) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
      7. \( \alpha ^3 + \beta ^3\)
        because \( \alpha ^3 + \beta ^3 =(\alpha + \beta)^3-3\alpha \beta (\alpha + \beta)\)
        means \( \alpha ^3 + \beta ^3\) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
      8. \( \alpha ^3 - \beta ^3\)
        because \( \alpha ^3 - \beta ^3 =\sqrt{(\alpha + \beta)^2-4\alpha \beta } ( \alpha ^2 + \alpha \beta + \beta ^2) \)
        means \( \alpha ^3 - \beta ^3 \) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
      9. \( (\alpha - \beta )^3\)
        because \( (\alpha - \beta )^3 = \left ( \sqrt{(\alpha + \beta)^2-4\alpha \beta } \right )^3\)
        means \( (\alpha - \beta )^3 \) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
      10. \( \alpha ^4 - \beta ^4\)
        because \( \alpha ^4 - \beta ^4 = (\alpha ^ 2+ \beta ^2)(\alpha + \beta) \sqrt{(\alpha + \beta)^2-4\alpha \beta }\)
        means \( \alpha ^4 - \beta ^4 \) can be expressed in terms of \( \alpha + \beta\) or \( \alpha \beta\)
      11. \( \frac{1}{\beta }+ \frac{1}{\alpha }\)
        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 \( \alpha \beta\)
      12. \(\frac{\alpha}{\beta }+ \frac{\beta}{\alpha } \)
        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 \( \alpha \beta\)
      13. \(\frac{\alpha ^2}{\beta }+ \frac{\beta ^2}{\alpha } \)
        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

      Common Root

      1. 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\( \frac{ \alpha ^2}{b_1c_2-b_2c_1} =\frac{ \alpha}{a_2c_1-a_1c_2} =\frac{1}{a_1b_2-a_2b_1}\)
        or \(\alpha= \frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1} =\frac{b_1c_2-b_2c_1}{a_2c_1-a_1c_2} \)
        or \( (a_2c_1-a_1c_2)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1) \)
        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
        \( (a_2c_1-a_1c_2)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1) \)

      2. 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]

      1. Determine the nature of the roots of each of the following equations
        1. \( x^2-6x+5=0 \)

          Solution 👉 Click Here

        2. \( x^2-4x-3=0 \)

          Solution 👉 Click Here

        3. \( x^2-6x+9=0 \)

          Solution 👉 Click Here

        4. \( 4x^2-4x+1=0 \)

          Solution 👉 Click Here

        5. \( 2x^2-9x+35=0 \)

          Solution 👉 Click Here

        6. \( 4x^2+8x-5=0 \)

          Solution 👉 Click Here

      2. For what values of p will the equation \( 5x^2-px+45=0\)?

        Solution 👉 Click Here

      3. if the equation \( x^2+2(k+2)x+9k=0\) has equal roots, find k.

        Solution 👉 Click Here

      4. For what value of a will the equation \(x^2-(3a-1)x+2(a^2-1)=0 \) have equal roots?

        Solution 👉 Click Here

      5. If the roots of the equation \((a^2+b^2)x^2-2(ac+bd)x+(c^2+d^2)=0 \) are equal, then show that \(\frac{a}{b}=\frac{c}{d}\)

        Solution 👉 Click Here

      6. Show that the roots of the equation \( (a^2-bc)x^2+2(b^2-ca)x+(c^2-ab)=0\) will be equal, if either \(b=0\) or \(a^3+b^3+c^3-3abc=0\)

        Solution 👉 Click Here

      7. If \(a,b,c\) are rational and \( a+b+c=0\), show that the roots \((b+c-a)x^2+(c+a-b)x+(a+b-c)=0\) are rational.

        Solution 👉 Click Here

      8. Prove that the roots of the equation \((x-a)(x-b)=k^2\) are real for all values of k.

        Solution 👉 Click Here

      9. Show that the roots of the equation \(x^2-4abx+(a^2+2b^2)^2=0 \) are imaginary.

        Solution 👉 Click Here

      10. If the roots of the quadratic equation \(qx^2+2px+2q=0\) are real and unequal, prove that the roots of the equation \((p+q)x^2+2qx+(p-q)=0\) are imaginary

        Solution 👉 Click Here

      Grade 11 Mathematics: Quadratic Equation [BCB Exercise 6.2]

      1. From the equation whose roots are
        1. 3,-2

          Solution 👉 Click Here

        2. -5,4

          Solution 👉 Click Here

        3. \(\sqrt{3},-\sqrt{3}\)

          Solution 👉 Click Here

        4. \(\frac{1}{2} (-1+\sqrt{5}),\frac{1}{2} (-1-\sqrt{5}) \)

          Solution 👉 Click Here

        5. -3+5i,-3-5i

          Solution 👉 Click Here

        6. a+ib,a-ib

          Solution 👉 Click Here

        1. Find a quadratic equation whose roots are twice the roots of \(4x^2+8x-5=0\)

          Solution 👉 Click Here

        2. Find a quadratic equation whose roots are reciprocals of the roots of \(3x^2-5x-2=0\)

          Solution 👉 Click Here

        3. Find a quadratic equation whose roots are greater by h than the roots of \(x^2-px+q=0\)

          Solution 👉 Click Here

        4. Find a quadratic equation whose roots are the squares of the roots of \(3x^2-5x-2=0\)

          Solution 👉 Click Here

      3. Find a quadratic equation with rational coefficients one of whose roots is
        1. 4+3i

          Solution 👉 Click Here

        2. \(\frac{1}{5+3i}\)

          Solution 👉 Click Here

        3. \(2+\sqrt{3}\)

          Solution 👉 Click Here

      4. Find the value of k so that the equation
        1. \(2x^2+kx-15=0\) has one root 3

          Solution 👉 Click Here

        2. \(3x^2+kx-2=0\) has roots whose sum is equal to 6

          Solution 👉 Click Here

        3. \(2x^2+(4-k)x-17=0\) has roots equal but opposite in sign

          Solution 👉 Click Here

        4. \(3x^2+(5+k)x+8=0\) has roots numerically equal but opposite in sign

          Solution 👉 Click Here

        5. \(3x^2+7x+6-k=0\) has one root equal to zero

          Solution 👉 Click Here

        6. \(4x^2-17x+k=0\) has the reciprocal roots

          Solution 👉 Click Here

        7. \(4x^2+kx+5=0\) has roots whose difference is \(\frac{1}{4}\)

          Solution 👉 Click Here

      5. Show that -1 is a root of the equation \((a+b-2c)x^2+(2a-b-c)x+(c+a-2b)=0\). Find the other root.

        Solution 👉 Click Here

      6. 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.

        Solution 👉 Click Here

      7. If the roots of the equation \(x^2+ax+c=0\) differ by 1, prove that \(a^2=4c+1\)

        Solution 👉 Click Here

      8. If \( \alpha, \beta\) are the roots of the equation \(x^2-x-6=0\), find the equation whose roots are
        1. \(\alpha ^2 \beta ^{-1}\)and \(\beta ^2 \alpha ^{-1}\)

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        2. \(\alpha + \frac{1}{\beta}\) and \(\beta + \frac{1}{\alpha}\)

          Solution 👉 Click Here

      9. If \( \alpha, \beta\) are the roots of the equation \(ax^2+bx+c=0\), find the equation whose roots are
        1. \(\alpha \beta ^{-1}\)and \(\beta \alpha ^{-1}\)

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        2. \(\alpha ^3 \) and \(\beta ^3\)

          Solution 👉 Click Here

        3. \((\alpha-\beta)^2 \) and \((\alpha+\beta)^2 \)

          Solution 👉 Click Here

        4. the reciprocal of the roots of given equation

          Solution 👉 Click Here

        1. 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

        2. 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

        3. For what values of m, the equation \(x^2-mx+m+1=0\) may have its root in the ratio 2:3

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        1. If \( \alpha, \beta\) are the roots of the equation \(px^2+qx+q=0\), prove that \(\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{q}{p}}=0\)

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        2. 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

      12. 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

      Grade 11 Mathematics: Quadratic Equation [BCB Exercise 6.3]

      1. Show that each pair of following equations has a common root
        1. \( x^2-8x+15=0\) and \(2x^2-x-15 =0\)

          Solution 👉 Click Here

        2. \( 3x^2-8x+4=0\) and \(4x^2-7x-2 =0\)

          Solution 👉 Click Here

      2. Find the value of p so that each pair of the equations may have one root common
        1. \( 4x^2+px-12=0\) and \(4x^2+3px-4 =0\)

          Solution 👉 Click Here

        2. \( 2x^2+px-1=0\) and \(3x^2-2x-5 =0\)

          Solution 👉 Click Here

      3. 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 \(\frac{pq'-p'q}{q-q'} \) or \(\frac{q-q'}{p'-p} \)

        Solution 👉 Click Here

      4. 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

      5. 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

      6. 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

      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 =\(\alpha ^2 \beta ^{-1}+ \beta ^2 \alpha ^{-1}\)
      =\( \frac{\alpha ^2}{\beta}+ \frac{\beta ^2}{\alpha}\)
      =\( \frac{\alpha ^3+\beta ^3}{\alpha \beta}\)
      =\( \frac{(\alpha +\beta) ^3-3\alpha \beta (\alpha +\beta) }{\alpha \beta}\)
      =\( \frac{(\alpha +\beta) ^3} {\alpha \beta} - \frac{3\alpha \beta (\alpha +\beta) }{\alpha \beta}\)
      =\( \frac{(\alpha +\beta) ^3} {\alpha \beta} - 3 (\alpha +\beta) \)
      =\( \frac{\left (-\frac{b}{a} \right ) ^3} {\frac{c}{a}} - 3 \left (-\frac{b}{a} \right ) \)
      =\( \frac{-b^3}{ca^2} + \frac{3b}{a} \)
      =\( \frac{3abc-b^3}{a^2c} \)
      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 \( x^2- \left ( \frac{3abc-b^3}{a^2c} \right) x + \left ( \frac{c}{a} \right)=0\)
      or \( a^2cx^2+ (b^3-3abc)x + ac^2=0\)
      This is the required quadratic equation

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