Question

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

• A. $(\sqrt{a}-\sqrt{b}{)}^2$
• B. $(\sqrt{a}+\sqrt{b}{)}^2$
• C. $a+b$
• D. $a-b$

Answer 1

Answer: (A), (B)

The correct options are
A $(\sqrt{a}+\sqrt{b}{)}^2$
B $(\sqrt{a}-\sqrt{b}{)}^2$

Given, $\frac{p}{2x}=\frac{a}{x+c}+\frac{b}{x-c}$
$\Rightarrow \frac{p(x^2-c^2)}{2x}=a(x-c)+b(x+c)$
$\Rightarrow \frac{p(x^2-c^2)}{2x}=(ax+bx)+bc-ac$
$\Rightarrow p(x^2-c^2)=2(a+b)x^2+2(bc-ac)x\Rightarrow (p-2a-2b)x^2-2(bc-ac)x-{pc}^2=0$
For equal roots
$D=0\Rightarrow 4{(bc-ac)}^2-4(p-2a-2b)(-{pc}^2)=0\Rightarrow {(bc-ac)}^2+p{c}^2(p-2a-2b)=0$
$\Rightarrow p={(\sqrt{a}+\sqrt{b})}^2$ or $p={(\sqrt{a}-\sqrt{b})}^2$