First slide

Theory of equations

Question

If one root of the quadratic equation ${px}^{2} + qx + r = 0 (p \neq 0)$ is $\frac{\sqrt{a}}{\sqrt{a} + \sqrt{a - b}}$ , where $p, q, r \in Q$ , and, a, b are positive integers which are not perfect square of any integer, then the other root is

Moderate

A

$\frac{\sqrt{a}}{\sqrt{a} - \sqrt{a - b}}$
B

$\frac{\sqrt{a} - \sqrt{a - b}}{\sqrt{b}}$
C

$a + \frac{\sqrt{a (a - b)}}{b}$
D

$\frac{a + \sqrt{a (a - b)}}{b}$

Solution

One root $\frac{\sqrt{a}}{\sqrt{a} + \sqrt{a - b}}$
On rationalizing, we get
$\begin{array}{l} \frac{\sqrt{a}}{\sqrt{a} + \sqrt{a - b}} \times \frac{\sqrt{a} - \sqrt{a - b}}{\sqrt{a} - \sqrt{a - b}} \\ = \frac{\sqrt{a} \cdot (\sqrt{a} - \sqrt{a - b})}{b} = \frac{a - \sqrt{a (a - b)}}{b} \end{array}$
$∴ Other root = \frac{a + \sqrt{a (a - b)}}{b}$
Rationalizing, we get
$\begin{array}{l} \frac{a + \sqrt{a (a - b)}}{b} \times \frac{a - \sqrt{a (a - b)}}{a - \sqrt{a (a - b)}} \\ = \frac{a^{2} - (a^{2} - ab)}{b (a - \sqrt{a (a - b)})} = \frac{\sqrt{a}}{\sqrt{a} - \sqrt{a - b}} \end{array}$

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