Precalculus Examples

$x^{4} - 8 x^{2} + 16$

Step 1

Set $x^{4} - 8 x^{2} + 16$ equal to $0$ .

$x^{4} - 8 x^{2} + 16 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

$u^{2} - 8 u + 16 = 0$

$u = x^{2}$

Step 2.2

Factor using the perfect square rule.

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Step 2.2.1

Rewrite $16$ as $4^{2}$ .

$u^{2} - 8 u + 4^{2} = 0$

Step 2.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 u = 2 \cdot u \cdot 4$

Step 2.2.3

Rewrite the polynomial.

$u^{2} - 2 \cdot u \cdot 4 + 4^{2} = 0$

Step 2.2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 4$ .

${(u - 4)}^{2} = 0$

Step 2.3

Set the $u - 4$ equal to $0$ .

$u - 4 = 0$

Step 2.4

Add $4$ to both sides of the equation.

$u = 4$

Step 2.5

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = 4$

Step 2.6

Solve the equation for $x$ .

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Step 2.6.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Step 2.6.2

Simplify $\pm \sqrt{4}$ .

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Step 2.6.2.1

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Step 2.6.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

Step 2.6.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.6.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Step 2.6.3.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

Step 2.6.3.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

Step 3