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 4 of 5: Bound the parameter

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In plain words

The upper bound on $x$ becomes the exact admissible interval for $p$ .

x\le\frac2{\sqrt3}\Longrightarrow (3p-4)(p+4)\le0\Longrightarrow p\le\frac43

Detailed analysis

Use the bound on $x$ in the formula for $x^2$ . After clearing the positive denominator and simplifying, the condition is $(3p-4)(p+4)\le0$ . Together with $p\ge0$ , this gives $0\le p\le4/3$ .