Question

If $c\ne 0$ and the equation $p\left(2x\right)=a(x+c)+b(x-c)$ 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 $(\sqrt{a}+\sqrt{b}{)}^2$

We can write the equation as

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

or$p(x^2-c^2)=2(a+b)x^2-2xc(a-b)$

or$(2a+2b-p)x^2-2c(a-b)x+p{c}^2=0$

For this equation to have equal roots,

$c^2{(a-b)}^2-p{c}^2(2a+2b-p)=0{(a-b)}^2-2p(a+b)+p^2=0\because c^2\ne 0{(p-(a+b\left)\right)}^2={(a+b)}^2-{(a-b)}^2{(p-(a+b\left)\right)}^2=4abp-(a+b)=\pm 2\sqrt{ab}p=a+b\pm 2\sqrt{ab}\therefore p={(\sqrt{a}\pm \sqrt{b})}^2$