Question

If $X$ denotes the set of real numbers $p$ for which the equation $x^2=p(x+p)$ has its roots greater than $p$, then $X$ is equal to

• A. $(-2,-\frac{1}{2})$
• B. $(-\frac{1}{2},\frac{1}{4})$
• C. Null set
• D. $(-\infty ,0)$

Answer 1

The correct option is C Null set

Given, $x^2=p(x+p)\Rightarrow x^2-px-p^2=0$
Now, given both roots are greater than $p$, so,
$f\left(p\right)>0$
$\Rightarrow p^2-p^2-p^2>0$
$\Rightarrow -p^2>0$
which is not possible so the solution set is a null set.
Hence, option 'C' is correct.