Find all real solutions of $\sqrt{x^2-p} + 2 \sqrt{x^2-1}=x$ for each value of $p$

Question

Find all the real solutions of the equation $$\sqrt{x^2-p} + 2 \sqrt{x^2-1}=x$$ for each real value of $p$.

I know that $\sqrt{r}$ means the non-negative square root of r', and r has to be non-negative and a real number for this to make sense. However, i'm not sure what they are asking.

I have tried to take p out of the equation several times and failed. Can someone please help?

Answer 1

You should show your attempts at the work but this may help

$$x = \pm \frac{ \sqrt{-p^2 + 8 p - 16}}{2 \sqrt{2} \sqrt{p - 2}}\quad\text{where}\quad p=4$$

Anything $<>4$ is negative under the numerator radical.

Answer 2

The equation imposes a few conditions:

1. $x^2\ge p$
2. $x^2\ge 1$
3. $x\ge0$

(the last one because the left-hand side is nonnegative). Now you can square both sides and set $t=x^2$, getting $$ t-p+4(t-1)+4\sqrt{(t-p)(t-1)}=t $$ that becomes $$ 4\sqrt{(t-p)(t-1)}=p+4-4t $$ This imposes the further condition that $4t\le p+4$ and you can square again.

Solve the resulting equation, apply the conditions and you'll find the required solutions (this might impose conditions on $p$ as well).

Answer 3

Square the equation

$$\sqrt{x^2-p} + 2 \sqrt{x^2-1}=x\tag 1$$

twice to get the nominal solution

$$x^2 = \frac {(4-p)^2}{8(2-p)}\tag 2$$

with $p<2$. To exclude spurious solutions from the square operation, substitute (2) into (1) to have

$$|3p-4| +2|p| = 4-p$$

which is valid only for $0\le p \le\frac43$. Thus, the solutions are

$$x = \frac {4-p}{2\sqrt{4-2p}}\>\>\>\>\text{for}\>\>\>\> 0\le p \le \frac43$$

and there are no solutions for other values of $p$.

Answer 4

In the value of $x^2=((4-p)^2)/(8(2-p)) $ given by @ Quanto above put $p=2(1-w^2)$ & we get

$x=(w^2+1)/(2w)$

Substituting value of this (x,p) back in the equation given by "OP" we get w=1. Hence $(x,p)=(1,0)$ is the only rational solution.