Solve (x)^(4)+(1)/((x)^(4))=47 | 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

$${ x }^{ 4 } + \frac{ 1 }{ { x }^{ 4 } } =47$$

Solve for x (complex solution)

$x\in \frac{-\sqrt{5}-3}{2},-\frac{\sqrt{5}i}{2}-\frac{3}{2}i,\frac{\sqrt{5}i}{2}+\frac{3}{2}i,\frac{\sqrt{5}+3}{2},\frac{\sqrt{5}i}{2}-\frac{3}{2}i,\frac{\sqrt{5}-3}{2},-\frac{\sqrt{5}i}{2}+\frac{3}{2}i,\frac{3-\sqrt{5}}{2}$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x^{4}$.

$$x^{4}x^{4}+1=47x^{4}$$

To multiply powers of the same base, add their exponents. Add $4$ and $4$ to get $8$.

$$x^{8}+1=47x^{4}$$

Subtract $47x^{4}$ from both sides.

$$x^{8}+1-47x^{4}=0$$

Substitute $t$ for $x^{4}$.

$$t^{2}-47t+1=0$$

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 $a$, $-47$ for $b$, and $1$ for $c$ in the quadratic formula.

$$t=\frac{-\left(-47\right)±\sqrt{\left(-47\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{47±21\sqrt{5}}{2}$$

Solve the equation $t=\frac{47±21\sqrt{5}}{2}$ when $±$ is plus and when $±$ is minus.

$$t=\frac{21\sqrt{5}+47}{2}$$ $$t=\frac{47-21\sqrt{5}}{2}$$

Since $x=t^{4}$, the solutions are obtained by solving the equation for each $t$.

$$x=-\left(\frac{\sqrt{5}i}{2}+\frac{3}{2}i\right)$$ $$x=-\frac{\sqrt{5}+3}{2}$$ $$x=\frac{\sqrt{5}i}{2}+\frac{3}{2}i$$ $$x=\frac{\sqrt{5}+3}{2}$$ $$x=-\left(-\frac{\sqrt{5}i}{2}+\frac{3}{2}i\right)$$ $$x=-\frac{3-\sqrt{5}}{2}$$ $$x=-\frac{\sqrt{5}i}{2}+\frac{3}{2}i$$ $$x=\frac{3-\sqrt{5}}{2}$$

Solve for x

$x=\frac{-\sqrt{5}-3}{2}\approx -2.618033989$
$x = \frac{\sqrt{5} + 3}{2} \approx 2.618033989$
$x=\frac{3-\sqrt{5}}{2}\approx 0.381966011$
$x=\frac{\sqrt{5}-3}{2}\approx -0.381966011$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x^{4}$.

$$x^{4}x^{4}+1=47x^{4}$$

To multiply powers of the same base, add their exponents. Add $4$ and $4$ to get $8$.

$$x^{8}+1=47x^{4}$$

Subtract $47x^{4}$ from both sides.

$$x^{8}+1-47x^{4}=0$$

Substitute $t$ for $x^{4}$.

$$t^{2}-47t+1=0$$

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 $a$, $-47$ for $b$, and $1$ for $c$ in the quadratic formula.

$$t=\frac{-\left(-47\right)±\sqrt{\left(-47\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{47±21\sqrt{5}}{2}$$

Solve the equation $t=\frac{47±21\sqrt{5}}{2}$ when $±$ is plus and when $±$ is minus.

$$t=\frac{21\sqrt{5}+47}{2}$$ $$t=\frac{47-21\sqrt{5}}{2}$$

Since $x=t^{4}$, the solutions are obtained by evaluating $x=±\sqrt[4]{t}$ for positive $t$.

$$x=\frac{\sqrt{5}+3}{2}$$ $$x=-\frac{\sqrt{5}+3}{2}$$ $$x=\frac{3-\sqrt{5}}{2}$$ $$x=-\frac{3-\sqrt{5}}{2}$$