Problem 1

For which real values of

p

does the equation

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

have real roots? What are the roots?

Step 1 of 5: Restrict the domain

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In plain words

The radicals force the solution into a short positive interval before any squaring is done.

x\ge 0,\qquad 2\sqrt{x^2-1}\le x,\qquad \sqrt{x^2-p}\le x

Detailed analysis

The left side is nonnegative, so $x\ge0$ . Since $2\sqrt{x^2-1}\le x$ , we have $x\le2/\sqrt3$ ; also $x^2\ge1$ , hence $x\ge1$ . Finally $\sqrt{x^2-p}\le x$ implies $p\ge0$ .