Question

If the ratio of the roots of equation $p{x}^2+qx+q=0$ is $a:b$, then$\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}+\sqrt{\frac{2q}{p}}=0$.

• A. True
• B. False

Answer 1

The correct option is B False

Let the roots of the equation $p{x}^2+qx+q=0$ be $ax$ and $bx$

then sum of roots=$ax+bx=\frac{-q}{p}-\left(1\right)$

product of roots $ax\times bx=\frac{q}{p}-\left(2\right)$

From eqs,(1) and (2) we have

$x(a+b)=-x^{2}ab$

$x[xab+(a+b\left)\right]=0$

$⟹x=-\frac{a+b}{ab}$

Therefore the roots are $ax=\frac{-(a+b)}{b}$ and $bx=\frac{-(a+b)}{a}$

$\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}+\sqrt{\frac{2p}{q}}=0⟹\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}+\sqrt{\frac{2(a+b{)}^2}{ab}}\ne 0$

Given statement is false.