Algebra Examples

$| x^{2} - 2 x | = 35$

Step 1

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

$x^{2} - 2 x = \pm 35$

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.

$x^{2} - 2 x = 35$

Step 2.2

Subtract $35$ from both sides of the equation.

$x^{2} - 2 x - 35 = 0$

Step 2.3

Factor $x^{2} - 2 x - 35$ using the AC method.

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Step 2.3.1

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 $- 35$ and whose sum is $- 2$ .

$- 7, 5$

Step 2.3.2

Write the factored form using these integers.

$(x - 7) (x + 5) = 0$

Step 2.4

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

$x - 7 = 0$

$x + 5 = 0$

Step 2.5

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 2.5.2

Add $7$ to both sides of the equation.

$x = 7$

Step 2.6

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 2.6.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.7

The final solution is all the values that make $(x - 7) (x + 5) = 0$ true.

$x = 7, - 5$

Step 2.8

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

$x^{2} - 2 x = - 35$

Step 2.9

Add $35$ to both sides of the equation.

$x^{2} - 2 x + 35 = 0$

Step 2.10

Use the quadratic formula to find the solutions.

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

Step 2.11

Substitute the values $a = 1$ , $b = - 2$ , and $c = 35$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot 35)}}{2 \cdot 1}$

Step 2.12

Simplify.

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Step 2.12.1

Simplify the numerator.

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Step 2.12.1.1

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 35}}{2 \cdot 1}$

Step 2.12.1.2

Multiply $- 4 \cdot 1 \cdot 35$ .

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Step 2.12.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 35}}{2 \cdot 1}$

Step 2.12.1.2.2

Multiply $- 4$ by $35$ .

$x = \frac{2 \pm \sqrt{4 - 140}}{2 \cdot 1}$

Step 2.12.1.3

Subtract $140$ from $4$ .

$x = \frac{2 \pm \sqrt{- 136}}{2 \cdot 1}$

Step 2.12.1.4

Rewrite $- 136$ as $- 1 (136)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 136}}{2 \cdot 1}$

Step 2.12.1.5

Rewrite $\sqrt{- 1 (136)}$ as $\sqrt{- 1} \cdot \sqrt{136}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{136}}{2 \cdot 1}$

Step 2.12.1.6

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

$x = \frac{2 \pm i \cdot \sqrt{136}}{2 \cdot 1}$

Step 2.12.1.7

Rewrite $136$ as $2^{2} \cdot 34$ .

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Step 2.12.1.7.1

Factor $4$ out of $136$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (34)}}{2 \cdot 1}$

Step 2.12.1.7.2

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

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 34}}{2 \cdot 1}$

Step 2.12.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{34})}{2 \cdot 1}$

Step 2.12.1.9

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{34}}{2 \cdot 1}$

Step 2.12.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 i \sqrt{34}}{2}$

Step 2.12.3

Simplify $\frac{2 \pm 2 i \sqrt{34}}{2}$ .

$x = 1 \pm i \sqrt{34}$

Step 2.13

The final answer is the combination of both solutions.

$x = 1 + i \sqrt{34}, 1 - i \sqrt{34}$

Step 2.14

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

$x = 7, - 5, 1 + i \sqrt{34}, 1 - i \sqrt{34}$