Exercise 466Algebra › Equations and radicalsHigh schoolAdded 3 Nov 2026

A simple radical equation

#radical equation#squaring#extraneous roots

Statement

Solve in $\mathbb{R}$ the equation
$\sqrt{2x+3}=x.$

Hint

Square both sides, but check that the solutions satisfy the original equation (the RHS must be $\ge0$ ).

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Condition. For $\sqrt{2x+3}$ to be real, $x\ge-3/2$ . Also, since $\sqrt{\cdot}\ge0$ , the RHS $x$ must be $\ge0$ .

Squaring. $2x+3=x^2$ , i.e. $x^2-2x-3=0$ .

Roots. $(x-3)(x+1)=0$ , so $x=3$ or $x=-1$ .

Check.
- $x=3$ : $\sqrt{9}=3$ , OK.
- $x=-1$ : RHS is $-1<0$ , but LHS $\sqrt1=1$ , so the original equation fails.

Conclusion. $x=3$ .