Algebra Examples

x^{2} - 4 x + 13 = 0

$x^{2} - 4 x + 13 = 0$

Step 1

Subtract

13

$13$ from both sides of the equation.

x^{2} - 4 x = - 13

$x^{2} - 4 x = - 13$

Step 2

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of

b

$b$ .

{(\frac{b}{2})}^{2} = {(- 2)}^{2}

${(\frac{b}{2})}^{2} = {(- 2)}^{2}$

Step 3

Add the term to each side of the equation.

x^{2} - 4 x + {(- 2)}^{2} = - 13 + {(- 2)}^{2}

$x^{2} - 4 x + {(- 2)}^{2} = - 13 + {(- 2)}^{2}$

Step 4

Simplify the equation.

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Step 4.1

Simplify the left side.

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Step 4.1.1

Raise

- 2

$- 2$ to the power of

2

$2$ .

x^{2} - 4 x + 4 = - 13 + {(- 2)}^{2}

$x^{2} - 4 x + 4 = - 13 + {(- 2)}^{2}$

x^{2} - 4 x + 4 = - 13 + {(- 2)}^{2}

$x^{2} - 4 x + 4 = - 13 + {(- 2)}^{2}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify

- 13 + {(- 2)}^{2}

$- 13 + {(- 2)}^{2}$ .

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Step 4.2.1.1

Raise

- 2

$- 2$ to the power of

2

$2$ .

x^{2} - 4 x + 4 = - 13 + 4

$x^{2} - 4 x + 4 = - 13 + 4$

Step 4.2.1.2

Add

- 13

$- 13$ and

4

$4$ .

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

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

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

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

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

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

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

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

Step 5

Factor the perfect trinomial square into

{(x - 2)}^{2}

${(x - 2)}^{2}$ .

{(x - 2)}^{2} = - 9

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

Step 6

Solve the equation for

x

$x$ .

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Step 6.1

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

x - 2 = \pm \sqrt{- 9}

$x - 2 = \pm \sqrt{- 9}$

Step 6.2

Simplify

\pm \sqrt{- 9}

$\pm \sqrt{- 9}$ .

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Step 6.2.1

Rewrite

- 9

$- 9$ as

- 1 (9)

$- 1 (9)$ .

x - 2 = \pm \sqrt{- 1 (9)}

$x - 2 = \pm \sqrt{- 1 (9)}$

Step 6.2.2

Rewrite

\sqrt{- 1 (9)}

$\sqrt{- 1 (9)}$ as

\sqrt{- 1} \cdot \sqrt{9}

$\sqrt{- 1} \cdot \sqrt{9}$ .

x - 2 = \pm \sqrt{- 1} \cdot \sqrt{9}

$x - 2 = \pm \sqrt{- 1} \cdot \sqrt{9}$

Step 6.2.3

Rewrite

\sqrt{- 1}

$\sqrt{- 1}$ as

i

$i$ .

x - 2 = \pm i \cdot \sqrt{9}

$x - 2 = \pm i \cdot \sqrt{9}$

Step 6.2.4

Rewrite

9

$9$ as

3^{2}

$3^{2}$ .

x - 2 = \pm i \cdot \sqrt{3^{2}}

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

Step 6.2.5

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

x - 2 = \pm i \cdot 3

$x - 2 = \pm i \cdot 3$

Step 6.2.6

Move

3

$3$ to the left of

i

$i$ .

x - 2 = \pm 3 i

$x - 2 = \pm 3 i$

x - 2 = \pm 3 i

$x - 2 = \pm 3 i$

Step 6.3

Move all terms not containing

x

$x$ to the right side of the equation.

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Step 6.3.1

Add

2

$2$ to both sides of the equation.

x = \pm 3 i + 2

$x = \pm 3 i + 2$

Step 6.3.2

Reorder

\pm 3 i

$\pm 3 i$ and

2

$2$ .

x = 2 \pm 3 i

$x = 2 \pm 3 i$

x = 2 \pm 3 i

$x = 2 \pm 3 i$

x = 2 \pm 3 i

$x = 2 \pm 3 i$