Question

If the ratio of the roots of $a{x}^2+2bx+c=0$ is same as the ratios of roots of $p{x}^2+2qx+r=0$, then

• A. $\frac{2b}{ac}=\frac{q^2}{pr}$
• B. $\frac{b}{ac}=\frac{q}{pr}$
• C. $\frac{b^2}{ac}=\frac{q^2}{pr}$
• D. $noneofthese$

Answer 1

The correct option is C $\frac{b^2}{ac}=\frac{q^2}{pr}$

Let ${\alpha }_1,{\beta }_1$ be the roots of First equation and ${\alpha }_2,{\beta }_2$ be the roots of second equation

Given that,

$\Rightarrow \frac{{\alpha }_1}{{\beta }_1}=\frac{{\alpha }_2}{{\beta }_2}$

Do componendo dividendo, we get

$\Rightarrow \frac{{\alpha }_1+{\beta }_1}{{\alpha }_1-{\beta }_1}=\frac{{\alpha }_2+{\beta }_2}{{\alpha }_2-{\beta }_2}$

Squaring on both sides

$\Rightarrow \frac{({\alpha }_1+{\beta }_1{)}^2}{({\alpha }_1-{\beta }_1{)}^2}=\frac{({\alpha }_2+{\beta }_2{)}^2}{({\alpha }_2-{\beta }_2{)}^2}$ ....$\left(1\right)$

We know that ${\alpha }_1+{\beta }_1=\frac{-b}{a}$

${\alpha }_1{\beta }_1=\frac{c}{a}$

$\Rightarrow ({\alpha }_1-{\beta }_1{)}^2=({\alpha }_1+{\beta }_1{)}^2-4{\alpha }_1{\beta }_1$

$\Rightarrow ({\alpha }_1-{\beta }_1{)}^2=(\frac{-b}{a}{)}^2-\frac{4c}{a}$

$\Rightarrow ({\alpha }_1-{\beta }_1{)}^2=\frac{b^2}{a^2}-\frac{4c}{a}$

Similarly, We know that ${\alpha }_2+{\beta }_2=\frac{-q}{p}$

${\alpha }_2{\beta }_2=\frac{r}{p}$

$\Rightarrow ({\alpha }_2-{\beta }_2{)}^2=\frac{q^2}{p^2}-\frac{4r}{p}$

Substitute these obtained values in $\left(1\right)$, we get,

$\Rightarrow \frac{\frac{b^2}{a^2}}{\frac{b^2}{a^2}-\frac{4c}{a}}=\frac{\frac{q^2}{p^2}}{\frac{q^2}{p^2}-\frac{4r}{p}}$

$\Rightarrow \frac{b^2}{b^2-4ac}=\frac{q^2}{q^2-4pr}$

$\Rightarrow b^2(q^2-4pr)=q^2(b^2-4ac)$

$\Rightarrow q^2{b}^2-4pr{b}^2=q^2{b}^2-4ac{q}^2$

$\Rightarrow 4pr{b}^2=4ac{q}^2$

$\Rightarrow pr{b}^2=ac{q}^2$

$\Rightarrow \frac{b^2}{ac}=\frac{q^2}{pr}$