Examples

2 (x^{2} - 1) = 16

$2 (x^{2} - 1) = 16$ ,

(0, 4)

$(0, 4)$

Step 1

Divide each term in

2 (x^{2} - 1) = 16

$2 (x^{2} - 1) = 16$ by

2

$2$ and simplify.

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Step 1.1

Divide each term in

2 (x^{2} - 1) = 16

$2 (x^{2} - 1) = 16$ by

2

$2$ .

\frac{2 (x^{2} - 1)}{2} = \frac{16}{2}

$\frac{2 (x^{2} - 1)}{2} = \frac{16}{2}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of

2

$2$ .

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Step 1.2.1.1

Cancel the common factor.

\frac{2 (x^{2} - 1)}{2} = \frac{16}{2}

$\frac{2 (x^{2} - 1)}{2} = \frac{16}{2}$

Step 1.2.1.2

Divide

x^{2} - 1

$x^{2} - 1$ by

1

$1$ .

x^{2} - 1 = \frac{16}{2}

$x^{2} - 1 = \frac{16}{2}$

x^{2} - 1 = \frac{16}{2}

$x^{2} - 1 = \frac{16}{2}$

x^{2} - 1 = \frac{16}{2}

$x^{2} - 1 = \frac{16}{2}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide

16

$16$ by

2

$2$ .

x^{2} - 1 = 8

$x^{2} - 1 = 8$

x^{2} - 1 = 8

$x^{2} - 1 = 8$

x^{2} - 1 = 8

$x^{2} - 1 = 8$

Step 2

Move all terms not containing

x

$x$ to the right side of the equation.

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Step 2.1

Add

1

$1$ to both sides of the equation.

x^{2} = 8 + 1

$x^{2} = 8 + 1$

Step 2.2

Add

8

$8$ and

1

$1$ .

x^{2} = 9

$x^{2} = 9$

x^{2} = 9

$x^{2} = 9$

Step 3

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

x = \pm \sqrt{9}

$x = \pm \sqrt{9}$

Step 4

Simplify

\pm \sqrt{9}

$\pm \sqrt{9}$ .

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Step 4.1

Rewrite

9

$9$ as

3^{2}

$3^{2}$ .

x = \pm \sqrt{3^{2}}

$x = \pm \sqrt{3^{2}}$

Step 4.2

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

x = \pm 3

$x = \pm 3$

x = \pm 3

$x = \pm 3$

Step 5

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

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Step 5.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

x = 3

$x = 3$

Step 5.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

x = - 3

$x = - 3$

Step 5.3

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

x = 3, - 3

$x = 3, - 3$

x = 3, - 3

$x = 3, - 3$

Step 6

Find the values of

n

$n$ that produce a value within the interval

(0, 4)

$(0, 4)$ .

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Step 6.1

The interval

(0, 4)

$(0, 4)$ does not contain

- 3

$- 3$ . It is not part of the final solution.

- 3

$- 3$ is not on the interval

Step 6.2

The interval

(0, 4)

$(0, 4)$ contains

3

$3$ .

x = 3

$x = 3$

x = 3

$x = 3$

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