Exercise 943Algebra › Equations and radicalsHigh schoolAdded 3 Nov 2026

A biquadratic equation

#biquadratic#substitution#real roots

Statement

Solve in $\mathbb{R}$ the equation
$x^4-5x^2+4=0.$

Hint

Set $y=x^2$ and solve the quadratic in $y$ .

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Substitution. Let $y=x^2\ge0$ . The equation becomes
$y^2-5y+4=0.$

Roots in $y$ . $(y-1)(y-4)=0$ , so $y=1$ or $y=4$ .

Back to $x$ . $x^2=1\Rightarrow x=\pm1$ ; $x^2=4\Rightarrow x=\pm2$ .

Conclusion. $x\in\{-2,-1,1,2\}$ .