Exercise 1677Algebra › Equations and radicalsL1Added 3 Nov 2026

A radical equation leading to a quartic

#radical#quartic#substitution#extraneous roots

Statement

Solve in $\mathbb{R}$ the equation
$\sqrt{x^2+3x+5}=2x-1.$

Hint

Square, then check that $2x-1\ge0$ .

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Condition. RHS must be $\ge0$ : $x\ge1/2$ .

Squaring. $x^2+3x+5=4x^2-4x+1$ , i.e. $3x^2-7x-4=0$ .

Roots. $\Delta=49+48=97$ , so $x=\frac{7\pm\sqrt{97}}{6}$ .

Check condition. $\sqrt{97}\approx9.85$ , so $x_1\approx2.81$ (valid) and $x_2\approx-0.475$ (rejected, $<1/2$ ).

Conclusion. $x=\frac{7+\sqrt{97}}{6}$ .