Question

The sum of integral values of $p$ for which the equation $\sqrt{x^2-p}+2\sqrt{x^2-1}=x$ has real solutions is

Answer 1

$\sqrt{x^2-p}+2\sqrt{x^2-1}=x\cdots \left(1\right)x^2-1\ge 0\Rightarrow x\ge 1\text{or}x\le -1$
And $x>0$
Therefore, $x\ge 1$
Squaring equation $\left(1\right)$, we get
$4\sqrt{(x^2-p)(x^2-1)}=p+4-4x^2$
So,
$p+4-4x^2\ge 0\Rightarrow p\ge 4(x^2-1)\Rightarrow p\ge 0$
Squaring again, we get
$\Rightarrow 16(x^4-x^2-p{x}^2+p)=(p+4{)}^2+16x^4-8x^2(p+4)$
$\Rightarrow 8x^2(2-p)=p^2-8p+16\Rightarrow 8x^2(2-p)=(p-4{)}^2$
So,
$2-p>0\Rightarrow p<2$
When $p=0$,
$x=1$
When $p=1$,
$3\sqrt{x^2-1}=x\Rightarrow x^2-1=\frac{x^2}{9}\Rightarrow x=\frac{3}{2\sqrt{2}}$
Sum of integral values of $p$ is $1$