MathLabs

Problem 1

For which real values of pp does the equation x2−p+2x2−1=x\sqrt{x^2-p}+2\sqrt{x^2-1}=x have real roots? What are the roots?
Step 2 of 5: Square once
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In plain words

Squaring removes one radical but records a sign condition that must be retained.

2x2−1=x−x2−p⟹2x2+p−4=2xx2−p2\sqrt{x^2-1}=x-\sqrt{x^2-p}\Longrightarrow 2x^2+p-4=2x\sqrt{x^2-p}
Detailed analysis

Isolate one radical and square. The unsquared equation requires the resulting right side 2x2+p−42x^2+p-4 to be nonnegative, so this operation introduces no accepted sign ambiguity when checked later.