Solve 6x ^ 4 + x ^ 2 - 1 = 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

$$6x^{4}+x^{2}-1=0$$

Solve for x (complex solution)

$x=\frac{\sqrt{3}}{3}\approx 0.577350269$
$x=-\frac{\sqrt{3}}{3}\approx -0.577350269$
$x=-\frac{\sqrt{2}i}{2}\approx -0-0.707106781i$
$x=\frac{\sqrt{2}i}{2}\approx 0.707106781i$

Steps Using the Quadratic Formula

Substitute $t$ for $x^{2}$.

$$6t^{2}+t-1=0$$

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 $6$ for $a$, $1$ for $b$, and $-1$ for $c$ in the quadratic formula.

$$t=\frac{-1±\sqrt{1^{2}-4\times 6\left(-1\right)}}{2\times 6}$$

Do the calculations.

$$t=\frac{-1±5}{12}$$

Solve the equation $t=\frac{-1±5}{12}$ when $±$ is plus and when $±$ is minus.

$$t=\frac{1}{3}$$ $$t=-\frac{1}{2}$$

Since $x=t^{2}$, the solutions are obtained by evaluating $x=±\sqrt{t}$ for each $t$.

$$x=-\frac{\sqrt{3}}{3}$$ $$x=\frac{\sqrt{3}}{3}$$ $$x=-\frac{\sqrt{2}i}{2}$$ $$x=\frac{\sqrt{2}i}{2}$$

Solve for x

$x=-\frac{\sqrt{3}}{3}\approx -0.577350269$
$x=\frac{\sqrt{3}}{3}\approx 0.577350269$

Steps Using the Quadratic Formula

Substitute $t$ for $x^{2}$.

$$6t^{2}+t-1=0$$

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 $6$ for $a$, $1$ for $b$, and $-1$ for $c$ in the quadratic formula.

$$t=\frac{-1±\sqrt{1^{2}-4\times 6\left(-1\right)}}{2\times 6}$$

Do the calculations.

$$t=\frac{-1±5}{12}$$

Solve the equation $t=\frac{-1±5}{12}$ when $±$ is plus and when $±$ is minus.

$$t=\frac{1}{3}$$ $$t=-\frac{1}{2}$$

Since $x=t^{2}$, the solutions are obtained by evaluating $x=±\sqrt{t}$ for positive $t$.

$$x=\frac{\sqrt{3}}{3}$$ $$x=-\frac{\sqrt{3}}{3}$$