Algebra Examples

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

Step 1

Add $16$ to both sides of the equation.

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

Step 2

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

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

Step 3

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

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

Step 4

Factor using the perfect square rule.

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Step 4.1

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

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

Step 4.2

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 4.3

Rewrite the polynomial.

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

Step 4.4

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 5

Replace all occurrences of $u$ with $x^{2}$ .

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

Step 6

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

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

Step 7

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

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

Step 8

Apply the product rule to $(x + 2) (x - 2)$ .

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

Step 9

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

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

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

Step 10

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

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Step 10.1

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

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

Step 10.2

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

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Step 10.2.1

Set the $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 10.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 11

Set ${(x - 2)}^{2}$ equal to $0$ and solve for $x$ .

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Step 11.1

Set ${(x - 2)}^{2}$ equal to $0$ .

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

Step 11.2

Solve ${(x - 2)}^{2} = 0$ for $x$ .

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Step 11.2.1

Set the $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 11.2.2

Add $2$ to both sides of the equation.

$x = 2$

Step 12

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

$x = - 2, 2$