Irrational equation

Let's check the conditions of existence for $x$ :

I) $x \geq 0$ , since it's a sum of two radicals.

II) $x^{2}-1 \geq 0$ , since $0 \leq x \Rightarrow 1 \leq x$ . Therefore we can make the following substitution: $x=\sec{\theta}, \ \theta \in \left[0, \frac{\pi}{2}\right).$

We got then the following equation to solve: $\sqrt{\sec^{2}{\theta}-p}+2\sqrt{\sec^{2}{\theta}-1}=\sec{\theta}$ $\Rightarrow \sqrt{\sec^{2}{\theta}-p}=\sec{\theta}-2\tan{\theta}.$

III) $\sec{\theta}-2\tan{\theta} \geq 0$ , since this expression equals to a radical.

$\therefore \ \sin{\theta} \leq \frac{1}{2} \Rightarrow 0 \leq \theta \leq \frac{\pi}{6}.$ $\sec^{2}{\theta}-p=\sec^{2}{\theta}-4\sec{\theta}\tan{\theta}+4\tan^{2}{\theta}$ $p=4\sec{\theta}\tan{\theta}-4\tan^{2}{\theta}$ $p=\frac{4\sin{\theta}\left(1-\sin{\theta}\right)}{cos^{2}{\theta}}$ $p=\frac{4\sin{\theta}}{1+\sin{\theta}}\Rightarrow p=4-\frac{4}{1+\sin{\theta}}$ $\Rightarrow \ max(p)=4-\frac{4}{1+\frac{1}{2}}=\frac{4}{3},$ which happens for $\theta=\frac{\pi}{6}.$ Hence: $x^{2}=\sec^{2}{\frac{\pi}{6}} \Rightarrow x^{2}=\frac{4}{3}.$ The final answer is $4+3=\boxed{7}.$

First we'll solve the equation for $x^2$ in terms of $p$ . Next, we'll take into consideration the constraints ( $x^2-p$ and $x^2-1$ both have to be non-negative in order for the equation to have a solution). Using those restraints, we can reduce the equation to something much easier to solve.

To solve the original equation, we move the $\sqrt{x^2-p}$ to the other side and square both sides, getting

$4(x^2-1) = x^2 -2x\sqrt{x^2-p} + x^2 -p$

After some simplification, the resulting equation is

$2x^2 + (p-4) = -2x \sqrt{x^2-p}$

We can then square both sides again, yielding

$4x^4 + 2(2x^2)(p-4)+(p-4)^2 = 4x^2 (x^2-p)$

By combining like terms, we can get

$8(p-2)x^2 + (p-4)^2 = 0$

So,

$\boxed{x^2 = \frac{-(p-2)(p+2)^2}{8}}$

Now we have to make sure that $x^2-1\geq 0$ .

$\frac{-(p-2)(p+2)^2}{8} - 1\geq 0$

$p(p^2+2p-4)\leq 0$

The maximum $p$ that satisfies the inequality is $-1+\sqrt{5}$ , but we realize that for $p = -1+\sqrt{5}$ , $x^2=1$ , which does not satisfy the $x^2-p\geq 0$ condition. Thus the more stringent constraint to be satisfied is $x^2-p\geq 0$ .

But since $x^2$ decreases as $p$ increases, the maximum value of $p$ has to happen when $x^2 = p$ , so we can set the left radical to 0, in which case the equation becomes $2\sqrt{x^2-1} = x$ .

$4(x^2-1) = x^2$

$3x^2 = 4$

$x^2 = \frac{4}{3}$

$\boxed{4+3=7}$