First slide

Theory of equations

Question

lf $c \neq 0$ and the equation $\frac{p}{(2 x)} = \frac{a}{(x + c)} + \frac{b}{(x - c)}$ has two equal roots, then p can be

Moderate

A

$(\sqrt{a} - \sqrt{b})^{2}$
B

$(\sqrt{a} + \sqrt{b})^{2}$
C

a+b
D

a-b

Solution

We can write the given equation as
$\begin{aligned} \frac{p}{2 x} = \frac{(a + b) x + c (b - a)}{x^{2} - c^{2}} \\ or & p (x^{2} - c^{2}) = 2 (a + b) x^{2} - 2 c (a - b) x \\ or & (2 a + 2 b - p) x^{2} - 2 c (a - b) x + {pc}^{2} = 0 \end{aligned}$
For this equation to have equal roots,
$\begin{aligned} c^{2} (a - b)^{2} - {pc}^{2} (2 a + 2 b - p) = 0 \\ or & (a - b)^{2} - 2 p (a + b) + p^{2} = 0 \\ or & [p - (a + b)]^{2} = (a + b)^{2} - (a - b)^{2} = 4 ab \\ or & p - (a + b) = \pm 2 \sqrt{ab} \\ or & p = a + b \pm 2 \sqrt{ab} = (\sqrt{a} \pm \sqrt{b})^{2} \end{aligned} [∵ c^{2} \neq 0]$

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