Question

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

• A. $a+b$
• B. $a-b$
• C. ${(\sqrt{a}\pm \sqrt{b})}^2$
• D. None of these

Answer 1

The correct option is C

${(\sqrt{a}\pm \sqrt{b})}^2$

Explantion for the correct options:

Find the value of $p$:

Given, $\frac{p}{2x}=\frac{a}{\left(x+c\right)}+\frac{b}{\left(x-c\right)}$

$\Rightarrow$ $\frac{p}{2x}=\frac{a\left(x-c\right)+b\left(x+c\right)}{\left(x+c\right)\left(x-c\right)}$

$\Rightarrow$ $\frac{p}{2x}=\frac{\left[ax-ac+bx+bc\right]}{\left[x^2-c^2\right]}$

$\Rightarrow$ $p\left(x^2-c^2\right)=2x\left[ax-ac+bx+bc\right]$

$\Rightarrow$ $p{x}^2-p{c}^2=2a{x}^2-2acx+2b{x}^2+2bcx$

$\Rightarrow$$p{x}^2-p{c}^2-2a{x}^2+2acx-2b{x}^2-2bcx=0$

$\Rightarrow$$p{x}^2-2a{x}^2-2b{x}^2+2acx-2bcx-p{c}^2=0$

$\Rightarrow$ $\left(p-2a-2b\right)x^2+2\left(ac-bc\right)x-p{c}^2=0$

For equal roots, $4{\left(ac-bc\right)}^2+4p{c}^2\left(p-2a-2b\right)=0$

${\left(a-b\right)}^2+p^2-2p\left(a+b\right)=0$

$\begin{array}{rcl}\left[p-{\left(a+b\right)}^2\right]& =& {\left(a+b\right)}^2-{\left(a-b\right)}^2\\ & =& 4ab\end{array}$

$\begin{array}{rcl}\therefore p& =& a+b\pm 2\sqrt{\mathrm{ab}}\\ & =& {\mathbf{(}\sqrt{\mathbf{a}}\mathbf{\pm }\sqrt{\mathbf{b}}\mathbf{)}}^{\mathbf{2}}\end{array}$

Hence, option(C) is correct.