Exercise 2132Algebra › Equations and radicalsL1Added 3 Nov 2026

A radical equation with parameter (advanced)

#parameter#radical equation#discussion

Statement

Find all real $a$ such that the equation
$\sqrt{x^2+ax+1}=x-1$
has a real solution.

Hint

Square carefully: the RHS must be $\ge0$ , and the resulting quadratic must have a root satisfying the condition.

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Solution

Condition. $x\ge1$ .

Squaring. $x^2+ax+1=x^2-2x+1$ , i.e. $ax=-2x$ , i.e. $(a+2)x=0$ .

Cases.
- If $a=-2$ : the equation is satisfied for all $x$ with $x\ge1$ . So any $x\ge1$ works. Also need $x^2+ax+1\ge0$ : $x^2-2x+1=(x-1)^2\ge0$ , always. So $a=-2$ works.
- If $a\neq-2$ : $x=0$ , but we need $x\ge1$ . Impossible.

Conclusion. The only value is $a=-2$ .

A radical equation with parameter (advanced)

Statement

Solution

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