Question

If the ratio of the roots of $x^2+bx+c=0$ and $x^2+qx+r=0$ is same, then

• A. $r^{2}c=q{b}^2$
• B. $r^{2}b=q{c}^2$
• C. $r{b}^2=c{q}^2$
• D. $r{c}^2=b{q}^2$

Answer 1

Answer: (C)

$r{b}^2=c{q}^2$

Given, ratio of roots of $x^2+bx+c=0$ and $x^2+qx+r=0$ is same.
Let ${\alpha }_1{\beta }_1$ be roots of $x^2+bx+c=0$
and ${\alpha }_2{\beta }_2$ be roots of $x^2+qx+r=0$
Given, $\frac{{\alpha }_1}{{\beta }_1}=\frac{{\alpha }_2}{{\beta }_2}$
$\Rightarrow \frac{{\alpha }_1+{\beta }_1}{{\alpha }_1-{\beta }_1}=\frac{{\alpha }_2+{\beta }_2}{{\alpha }_2-{\beta }_2}$
$\Rightarrow {\alpha }_1+{\beta }_1=\frac{-b}{a}=-b$ ....(Since $\Rightarrow a=1$ in this case)
$\Rightarrow {\alpha }_1{\beta }_1=c$
$\Rightarrow {\alpha }_1-{\beta }_1=\sqrt{b^2-4c}$ ....(Since $a=1$)
$\Rightarrow \frac{-b}{\sqrt{b^2-4c}}=\frac{-q}{\sqrt{q^2-4r}}$
Squaring on both sides, we get
$\Rightarrow b^2(q^2-4r)=q^2(b^2-4c)$
$\Rightarrow -4b^{2}r=-4c{q}^2$
$\Rightarrow r{b}^2=c{q}^2$