Question

If $a{x}^2+2bx+c=0$ and $a_1{x}^2+2b_{1}x+c_1=0$ have a common root and $\frac{a}{a_1},\frac{b}{b_1},\frac{c}{c_1}$ are in A.P, and ${b_1}^m=a_1{c}_1$.Find the value of $m$.

Answer 1

$a{x}^2+2bx+c=0$ and $a_1{x}^2+2b_{1}x+c_1=0$ have a common root $\frac{a}{a_1},\frac{b}{b_1},\frac{c}{c_1}$ are in A.P.

$(2b{c}_1-2b_{1}c)(2a{b}_1-2b{a}_1)=(-c_{1}a+c{a}_1{)}^2$

$4\left(c_1{b}_1\right)(\frac{b}{b_1}-\frac{c}{c_1})(\frac{a}{a_1}-\frac{b}{b_1})\left(a_1{b}_1\right)=\left(a_1{c}_1{)}^2\right(\frac{c}{c_1}-\frac{a}{a_1}{)}^2$

$4\left(b_1{)}^2\right(\frac{c}{c_1}-\frac{b}{b_1}\left)\right(\frac{b}{b_1}-\frac{a}{a_1})=(a_1{c}_1\left)\right(\frac{c}{c_1}-\frac{a}{a_1}{)}^2$ - Equation $1$

As, $\frac{a}{a_1},\frac{b}{b_1},\frac{c}{c_1}$ are in A.P., difference between them is same.

$\Rightarrow \frac{c}{c_1}-\frac{b}{b_1}=\frac{b}{b_1}-\frac{a}{a_1}$ - Equation 2

Difference between, $(\frac{c}{c_1}-\frac{a}{a_1})=2(\frac{c}{c_1}-\frac{b}{b_1})$ - Equation $3$

From Equation 3 & 2, Equation 1 becomes,

$4\left(b_1{)}^2\right(\frac{c}{c_1}-\frac{b}{b_1}\left)\right(\frac{c}{c_1}-\frac{b}{b_1})=(a_1{c}_1\left)\right(\frac{c}{c_1}-\frac{b}{b_1}{)}^2(2{)}^2$

$\Rightarrow b_1^2=a_1{c}_1$

$\Rightarrow m=2$