Question

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

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

Answer 1

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

Let $\alpha ,\beta$ be the roots of the equation are in ratio $m:n$, then $\alpha +\beta =\frac{-b}{a}$ and $\alpha \beta =\frac{c}{a}$

And $\frac{\alpha }{\beta }=\frac{m}{n}\Rightarrow \frac{\alpha }{m}=\frac{\beta }{n}=\frac{\alpha +\beta }{m+n}=k$

$\therefore \alpha +\beta =k(m+n),\alpha =km.\beta =kn$

Now ${(\alpha +\beta )}^2=k^2{(m+n)}^2$

$=\left(\frac{\alpha }{m}\right)\left(\frac{\beta }{n}\right){(m+n)}^2$

$\Rightarrow {\left(\frac{-b}{a}\right)}^{2}mn=\left(\frac{c}{a}\right){(m+n)}^2$

$\Rightarrow mn{b}^2={(m+n)}^{2}ac$ ...(1)

Also if the roots of equation $p{x}^2+2qx+r=0$ are also in the ration $m:n$, then

$mn{\left(2q\right)}^2={(m+n)}^{2}pr$ ...(2)

Dividing (1) and (2), we get

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