- Show that each pair of following equations has a common root
- \( x^2-8x+15=0\) and \(2x^2-x-15 =0\)
Solution 👉 Click Here
Solution :1(a)
Given quadratic equations are
\( x^2-8x+15=0\) and \(2x^2-x-15 =0\)
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
\( (a_2c_1-a_1c_2)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1) \)
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, \( x^2-8x+15=0\) and \(2x^2-x-15 =0\) has a common root.
- \( 3x^2-8x+4=0\) and \(4x^2-7x-2 =0\)
Solution 👉 Click Here
Solution :1(b)
Given quadratic equations are
\( 3x^2-8x+4=0\) and \(4x^2-7x-2 =0\)
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
\( (a_2c_1-a_1c_2)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1) \)
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, \( 3x^2-8x+4=0\) and \(4x^2-7x-2 =0\) has a common root.
- Find the value of p so that each pair of the equations may have one root common
- \( 4x^2+px-12=0\) and \(4x^2+3px-4 =0\)
Solution 👉 Click Here
Solution :2(a)
Given quadratic equations are
\( 4x^2+px-12=0\) and \(4x^2+3px-4 =0\)
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
\( (c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1) \)
or \( [(-12)(4)-(-4)(4)]^2 = [(4)(3p)-(4)(p)] [(p)(-4)-(3p)(-12)]\)
or \( [-48+16]^2 = [12p-4p] [-4p+36p]\)
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
- \( 2x^2+px-1=0\) and \(3x^2-2x-5 =0\)
Solution 👉 Click Here
Solution :2(b)
Given quadratic equations are
\(2x^2+px-1=0\) and \(3x^2-2x-5=0\)
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
\( (c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1) \)
or \( [(-1)(3)-(-5)(2)]^2 = [(2)(-2)-(3)(p)] [(p)(-5)-(-2)(-1)]\)
or \( [-3+10]^2 = [-4-3p] [-5p-2]\)
or \( [7]^2 = [4+3p] [2+5p]\)
or \(49 = 15p^2+26p+8\)
or \( 15p^2+26p-41=0\)
or \( 15p^2+(41-15)p-41=0\)
or \( 15p^2+41p-15p-41=0\)
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 \(\frac{pq'-p'q}{q-q'} \) or \(\frac{q-q'}{p'-p} \)
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
\((p-p') \alpha +(q-q')=0\)
or \( \alpha= \frac{q'-q}{p-p'} \)
or \( \alpha= \frac{q-q'}{p'-p} \)
Multiplying (1) by \(q'\), (2) by \(q\) and substracting, we get
\((q'-q) \alpha +(pq'-p'q)=0\)
or \( \alpha= \frac{pq'-p'q}{q-q'} \)
Thus,\(x^2+px+q=0\) and \(x^2+p'x+q'=0\) have one common root \(\alpha\) if \( \alpha= \frac{q-q'}{p'-p} \) or \( \alpha= \frac{pq'-p'q}{q-q'} \)
- 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
\( (c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1) \)
or \( [(q)(1)-(p)(1)]^2 = [(1)(q)-(1)(p)] [(p)(p)-(q)(q)]\)
or \( [q-p]^2 = [q-p] [p^2-q^2]\)
or \( (q-p) [(q-p)- (p^2-q^2)]=0\)
or \( q-p =0\) or \( p+q-1 =0\)
Thus,\(x^2+px+q=0\) and \(x^2+qx+p=0\) have one root common if \( p=q\) or \( p+q-1 =0\)
- 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
\( (c_1a_2-c_2a_1)^2 = (a_1b_2-a_2b_1) (b_1c_2-b_2c_1) \)
or \( [(c)(b)-(a)(a)]^2 = [(a)(c)-(b)(b)] [(b)(a)-(c)(c)]\)
or \( [bc-a^2]^2 = [ca-b^2] [ab-c^2]\)
or \( b^2c^2-2a^2bc+a^4=a^2bc-ac^3-ab^3+b^2c^2\)
or \( a^4=3a^2bc-ac^3-ab^3\)
or \( a^3=3abc-c^3-b^3\)
or \( a^3+b^3+c^3-3abc=0\)
or \( a^3+b^3+c^3-3abc=0\)
or \( (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\)
or Either \( (a+b+c)=0\) or \((a^2+b^2+c^2-ab-bc-ca)=0\)
or Either \( a+b+c=0\) or \(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
or Either \( a+b+c=0\) or \((a-b)^2+(b-c)^2+(c-a)^2=0\)
or Either \( a+b+c=0\) or \(a-b=0, b-c=0,c-a\)
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\(k=\frac{ab^2-ac^2}{ac-ab} \)
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\(k=\frac{ac-ab}{c-b} \)
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\( \alpha=-a-b\)
Next, suppose \(\beta \) be the other root of \(x^2+cx+ab=0\), then
\( k+\beta=-c\)
or\( a+\beta=-c\)
or\( \beta=-a-c\)
Now, new quadratic equation containing other roots \(\alpha, \beta\)is given by
\(x^2-(\alpha+ \beta)x+ (\alpha \beta)=0\)
or\(x^2-(-a-b-a-c)x+ (a-b)(a-c)=0\)
or\(x^2-(-2a-b-c)x+ (a^2-ab-ac+bc)=0\)
or\(x^2-(-2a+a)x+ (0+bc)=0\)
or\(x^2+ax+ bc=0\)
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