Question

If one root of the quadratic equation $p{x}^2+qx+r=0(p\ne 0)$ is a surd $\frac{\sqrt{a}}{\sqrt{a}+\sqrt{a-b}}$, where $p,q,r,a,b$ are all rationals, then the other root is

• A. $\frac{\sqrt{b}}{\sqrt{a}-\sqrt{a-b}}$
• B. $a+\frac{\sqrt{a(a-b)}}{b}$
• C. $\frac{a+\sqrt{a(a-b)}}{b}$
• D. $\frac{\sqrt{a}-\sqrt{a-b}}{\sqrt{b}}$

Answer 1

The correct option is C $\frac{a+\sqrt{a(a-b)}}{b}$

Given that one root of $p{x}^2+qx+r=0$ is $\frac{\sqrt{a}}{\sqrt{a}+\sqrt{a-b}}$ and $p,q,r$ are all rationals.
$\frac{\sqrt{a}}{\sqrt{a}+\sqrt{a-b}}$$=\frac{\sqrt{a}(\sqrt{a}-\sqrt{a-b})}{a-(a-b)}$
$=\frac{a-\sqrt{a(a-b)}}{b}$
As $p,q,r$ are rational, roots are conjugate surds.
$\therefore$ Other root is $\frac{a+\sqrt{a(a-b)}}{b}$
Hence, option C.