Question

If the ratio of the roots of the equation $a{x}^2+bx+c=0$ is $m:n$, then

• A. $\frac{m}{n}+\frac{n}{m}=\frac{b^2}{ac}$
• B. $\sqrt{\left(\frac{m}{n}\right)}+\sqrt{\left(\frac{n}{m}\right)}=\frac{\left|b\right|}{\sqrt{ca}}$
• C. $\sqrt{\left(\frac{m}{n}\right)}+\sqrt{\left(\frac{n}{m}\right)}=\frac{b^2}{ca}$
• D. $\sqrt{\left(\frac{m}{n}\right)}+\sqrt{\left(\frac{n}{m}\right)}=\frac{a^2}{b^2}$

Answer 1

The correct option is B $\sqrt{\left(\frac{m}{n}\right)}+\sqrt{\left(\frac{n}{m}\right)}=\frac{\left|b\right|}{\sqrt{ca}}$

Let $\alpha ,\beta$ be the roots of $a{x}^2+bx+c=0$

Then, $\alpha \beta =\frac{c}{a}$ and $\alpha +\beta =\frac{-b}{a}$

Now it is given that,

$\frac{\alpha }{\beta }=\frac{m}{n}$

Consider,

$\frac{m}{n}+\frac{n}{m}$

$=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$

$=\frac{{\alpha }^2+{\beta }^2}{\alpha .\beta }$

$=\frac{(\alpha +\beta {)}^2-2\alpha .\beta }{\alpha .\beta }$

$=\frac{\frac{b^2}{a^2}-\frac{c}{a}}{\frac{c}{a}}$

$=\frac{b^2-ac}{ac}$ ....... $\left(i\right)$

Similarly,

$\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}$

$=\sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}$

$=\frac{|\alpha +\beta |}{\sqrt{\alpha .\beta }}$

$=\frac{\left|\frac{-b}{a}\right|}{\sqrt{\frac{c}{a}}}$

$=\frac{|b|}{\sqrt{ac}}$