Algebra Examples

$| x - 9 | = x^{2} + 3$

Step 1

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x - 9 = \pm (x^{2} + 3)$

Step 2

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

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Step 2.1

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

$x - 9 = x^{2} + 3$

Step 2.2

Subtract $x^{2}$ from both sides of the equation.

$x - 9 - x^{2} = 3$

Step 2.3

Move all terms to the left side of the equation and simplify.

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Step 2.3.1

Subtract $3$ from both sides of the equation.

$x - 9 - x^{2} - 3 = 0$

Step 2.3.2

Subtract $3$ from $- 9$ .

$x - x^{2} - 12 = 0$

Step 2.4

Use the quadratic formula to find the solutions.

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

Step 2.5

Substitute the values $a = - 1$ , $b = 1$ , and $c = - 12$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (- 1 \cdot - 12)}}{2 \cdot - 1}$

Step 2.6

Simplify.

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Step 2.6.1

Simplify the numerator.

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Step 2.6.1.1

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 1 \cdot - 12}}{2 \cdot - 1}$

Step 2.6.1.2

Multiply $- 4 \cdot - 1 \cdot - 12$ .

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Step 2.6.1.2.1

Multiply $- 4$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{1 + 4 \cdot - 12}}{2 \cdot - 1}$

Step 2.6.1.2.2

Multiply $4$ by $- 12$ .

$x = \frac{- 1 \pm \sqrt{1 - 48}}{2 \cdot - 1}$

Step 2.6.1.3

Subtract $48$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 47}}{2 \cdot - 1}$

Step 2.6.1.4

Rewrite $- 47$ as $- 1 (47)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 47}}{2 \cdot - 1}$

Step 2.6.1.5

Rewrite $\sqrt{- 1 (47)}$ as $\sqrt{- 1} \cdot \sqrt{47}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{47}}{2 \cdot - 1}$

Step 2.6.1.6

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

$x = \frac{- 1 \pm i \sqrt{47}}{2 \cdot - 1}$

Step 2.6.2

Multiply $2$ by $- 1$ .

$x = \frac{- 1 \pm i \sqrt{47}}{- 2}$

Step 2.6.3

Simplify $\frac{- 1 \pm i \sqrt{47}}{- 2}$ .

$x = \frac{1 \pm i \sqrt{47}}{2}$

Step 2.7

The final answer is the combination of both solutions.

$x = \frac{1 + i \sqrt{47}}{2}, \frac{1 - i \sqrt{47}}{2}$

Step 2.8

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

$x - 9 = - (x^{2} + 3)$

Step 2.9

Simplify $- (x^{2} + 3)$ .

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Step 2.9.1

Rewrite.

$x - 9 = 0 + 0 - (x^{2} + 3)$

Step 2.9.2

Simplify by adding zeros.

$x - 9 = - (x^{2} + 3)$

Step 2.9.3

Apply the distributive property.

$x - 9 = - x^{2} - 1 \cdot 3$

Step 2.9.4

Multiply $- 1$ by $3$ .

$x - 9 = - x^{2} - 3$

Step 2.10

Add $x^{2}$ to both sides of the equation.

$x - 9 + x^{2} = - 3$

Step 2.11

Add $3$ to both sides of the equation.

$x - 9 + x^{2} + 3 = 0$

Step 2.12

Add $- 9$ and $3$ .

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

Step 2.13

Factor the left side of the equation.

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Step 2.13.1

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$u + u^{2} - 6 = 0$

Step 2.13.2

Factor $u + u^{2} - 6$ using the AC method.

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Step 2.13.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $1$ .

$- 2, 3$

Step 2.13.2.2

Write the factored form using these integers.

$(u - 2) (u + 3) = 0$

Step 2.13.3

Replace all occurrences of $u$ with $x$ .

$(x - 2) (x + 3) = 0$

Step 2.14

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

$x - 2 = 0$

$x + 3 = 0$

Step 2.15

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 2.15.1

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

$x - 2 = 0$

Step 2.15.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.16

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 2.16.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.16.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.17

The final solution is all the values that make $(x - 2) (x + 3) = 0$ true.

$x = 2, - 3$

Step 2.18

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

$x = \frac{1 + i \sqrt{47}}{2}, \frac{1 - i \sqrt{47}}{2}, 2, - 3$