Algebra Examples

$| x - 4 | = x^{2} - 6 x + 9$

Step 1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} - 6 x + 9 = | x - 4 |$

Step 2

Rewrite the equation as $| x - 4 | = x^{2} - 6 x + 9$ .

$| x - 4 | = x^{2} - 6 x + 9$

Step 3

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

$x - 4 = \pm (x^{2} - 6 x + 9)$

Step 4

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

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Step 4.1

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

$x - 4 = x^{2} - 6 x + 9$

Step 4.2

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} - 6 x + 9 = x - 4$

Step 4.3

Move all terms containing $x$ to the left side of the equation.

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Step 4.3.1

Subtract $x$ from both sides of the equation.

$x^{2} - 6 x + 9 - x = - 4$

Step 4.3.2

Subtract $x$ from $- 6 x$ .

$x^{2} - 7 x + 9 = - 4$

Step 4.4

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

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Step 4.4.1

Add $4$ to both sides of the equation.

$x^{2} - 7 x + 9 + 4 = 0$

Step 4.4.2

Add $9$ and $4$ .

$x^{2} - 7 x + 13 = 0$

Step 4.5

Use the quadratic formula to find the solutions.

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

Step 4.6

Substitute the values $a = 1$ , $b = - 7$ , and $c = 13$ into the quadratic formula and solve for $x$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (1 \cdot 13)}}{2 \cdot 1}$

Step 4.7

Simplify.

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Step 4.7.1

Simplify the numerator.

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Step 4.7.1.1

Raise $- 7$ to the power of $2$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot 1 \cdot 13}}{2 \cdot 1}$

Step 4.7.1.2

Multiply $- 4 \cdot 1 \cdot 13$ .

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Step 4.7.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot 13}}{2 \cdot 1}$

Step 4.7.1.2.2

Multiply $- 4$ by $13$ .

$x = \frac{7 \pm \sqrt{49 - 52}}{2 \cdot 1}$

Step 4.7.1.3

Subtract $52$ from $49$ .

$x = \frac{7 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 4.7.1.4

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

$x = \frac{7 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 4.7.1.5

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

$x = \frac{7 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 4.7.1.6

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

$x = \frac{7 \pm i \sqrt{3}}{2 \cdot 1}$

Step 4.7.2

Multiply $2$ by $1$ .

$x = \frac{7 \pm i \sqrt{3}}{2}$

Step 4.8

The final answer is the combination of both solutions.

$x = \frac{7 + i \sqrt{3}}{2}, \frac{7 - i \sqrt{3}}{2}$

Step 4.9

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

$x - 4 = - (x^{2} - 6 x + 9)$

Step 4.10

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- (x^{2} - 6 x + 9) = x - 4$

Step 4.11

Simplify $- (x^{2} - 6 x + 9)$ .

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Step 4.11.1

Rewrite.

$0 + 0 - (x^{2} - 6 x + 9) = x - 4$

Step 4.11.2

Simplify by adding zeros.

$- (x^{2} - 6 x + 9) = x - 4$

Step 4.11.3

Apply the distributive property.

$- x^{2} - (- 6 x) - 1 \cdot 9 = x - 4$

Step 4.11.4

Simplify.

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Step 4.11.4.1

Multiply $- 6$ by $- 1$ .

$- x^{2} + 6 x - 1 \cdot 9 = x - 4$

Step 4.11.4.2

Multiply $- 1$ by $9$ .

$- x^{2} + 6 x - 9 = x - 4$

Step 4.12

Move all terms containing $x$ to the left side of the equation.

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Step 4.12.1

Subtract $x$ from both sides of the equation.

$- x^{2} + 6 x - 9 - x = - 4$

Step 4.12.2

Subtract $x$ from $6 x$ .

$- x^{2} + 5 x - 9 = - 4$

Step 4.13

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

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Step 4.13.1

Add $4$ to both sides of the equation.

$- x^{2} + 5 x - 9 + 4 = 0$

Step 4.13.2

Add $- 9$ and $4$ .

$- x^{2} + 5 x - 5 = 0$

Step 4.14

Use the quadratic formula to find the solutions.

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

Step 4.15

Substitute the values $a = - 1$ , $b = 5$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (- 1 \cdot - 5)}}{2 \cdot - 1}$

Step 4.16

Simplify.

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Step 4.16.1

Simplify the numerator.

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Step 4.16.1.1

Raise $5$ to the power of $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 1 \cdot - 5}}{2 \cdot - 1}$

Step 4.16.1.2

Multiply $- 4 \cdot - 1 \cdot - 5$ .

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Step 4.16.1.2.1

Multiply $- 4$ by $- 1$ .

$x = \frac{- 5 \pm \sqrt{25 + 4 \cdot - 5}}{2 \cdot - 1}$

Step 4.16.1.2.2

Multiply $4$ by $- 5$ .

$x = \frac{- 5 \pm \sqrt{25 - 20}}{2 \cdot - 1}$

Step 4.16.1.3

Subtract $20$ from $25$ .

$x = \frac{- 5 \pm \sqrt{5}}{2 \cdot - 1}$

Step 4.16.2

Multiply $2$ by $- 1$ .

$x = \frac{- 5 \pm \sqrt{5}}{- 2}$

Step 4.16.3

Simplify $\frac{- 5 \pm \sqrt{5}}{- 2}$ .

$x = \frac{5 \pm \sqrt{5}}{2}$

Step 4.17

The final answer is the combination of both solutions.

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

Step 4.18

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

$x = \frac{7 + i \sqrt{3}}{2}, \frac{7 - i \sqrt{3}}{2}, \frac{5 + \sqrt{5}}{2}, \frac{5 - \sqrt{5}}{2}$