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 3 of 5: Square again and solve
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In plain words

The two radicals collapse to one rational expression for x2x^2.

(2x2+p−4)2=4x2(x2−p)⟹x2=(p−4)216−8p(2x^2+p-4)^2=4x^2(x^2-p)\Longrightarrow x^2=\frac{(p-4)^2}{16-8p}
Detailed analysis

Squaring the relation from the previous step and simplifying yields (16−8p)x2=(p−4)2(16-8p)x^2=(p-4)^2. Thus any solution must satisfy x2=(p−4)2/(16−8p)x^2=(p-4)^2/(16-8p).