Find the Roots (Zeros) (x^2+1)^2-x^2=0

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

Step 1

Factor the left side of the equation.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2} + 1$ and $b = x$ .

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

Simplify.

Reorder terms.

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

Reorder terms.

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

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

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

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

Step 3

Set $x^{2} + x + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + x + 1$ equal to $0$ .

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

Solve $x^{2} + x + 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Step 4

Set $x^{2} - x + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} - x + 1$ equal to $0$ .

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

Solve $x^{2} - x + 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

$x = \frac{1 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{1 \pm i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$x = \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

Step 5

The final solution is all the values that make $(x^{2} + x + 1) (x^{2} - x + 1) = 0$ true.

$x = - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

Step 6

Find the Roots (Zeros) (x^2+1)^2-x^2=0

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