Solve (x)^(4)-4(x)^(2)-x+2=0 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$${ x }^{ 4 } -4 { x }^{ 2 } -x+2=0$$

Solve for x

$x=\frac{-\sqrt{5}-1}{2}\approx -1.618033989$
$x=2$
$x=\frac{\sqrt{5}-1}{2}\approx 0.618033989$
$x=-1$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $2$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-1$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{4}-4x^{2}-x+2$ by $x+1$ to get $x^{3}-x^{2}-3x+2$. Solve the equation where the result equals to $0$.

$$x^{3}-x^{2}-3x+2=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $2$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=2$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}-x^{2}-3x+2$ by $x-2$ to get $x^{2}+x-1$. Solve the equation where the result equals to $0$.

$$x^{2}+x-1=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $1$ for $a$, $1$ for $b$, and $-1$ for $c$ in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\left(-1\right)}}{2}$$

Do the calculations.

$$x=\frac{-1±\sqrt{5}}{2}$$

Solve the equation $x^{2}+x-1=0$ when $±$ is plus and when $±$ is minus.

$$x=\frac{-\sqrt{5}-1}{2}$$ $$x=\frac{\sqrt{5}-1}{2}$$

List all found solutions.

$$x=-1$$ $$x=2$$ $$x=\frac{-\sqrt{5}-1}{2}$$ $$x=\frac{\sqrt{5}-1}{2}$$