Precalculus Examples

\sqrt{x + 20} = x

$\sqrt{x + 20} = x$

Step 1

To remove the radical on the left side of the equation, square both sides of the equation.

{\sqrt{x + 20}}^{2} = x^{2}

${\sqrt{x + 20}}^{2} = x^{2}$

Step 2

Simplify each side of the equation.

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Step 2.1

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt{x + 20}

$\sqrt{x + 20}$ as

{(x + 20)}^{\frac{1}{2}}

${(x + 20)}^{\frac{1}{2}}$ .

{({(x + 20)}^{\frac{1}{2}})}^{2} = x^{2}

${({(x + 20)}^{\frac{1}{2}})}^{2} = x^{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify

{({(x + 20)}^{\frac{1}{2}})}^{2}

${({(x + 20)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.1.1

Multiply the exponents in

{({(x + 20)}^{\frac{1}{2}})}^{2}

${({(x + 20)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.1.1.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

{(x + 20)}^{\frac{1}{2} \cdot 2} = x^{2}

${(x + 20)}^{\frac{1}{2} \cdot 2} = x^{2}$

Step 2.2.1.1.2

Cancel the common factor of

2

$2$ .

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Step 2.2.1.1.2.1

Cancel the common factor.

{(x + 20)}^{\frac{1}{2} \cdot 2} = x^{2}

${(x + 20)}^{\frac{1}{2} \cdot 2} = x^{2}$

Step 2.2.1.1.2.2

Rewrite the expression.

{(x + 20)}^{1} = x^{2}

${(x + 20)}^{1} = x^{2}$

{(x + 20)}^{1} = x^{2}

${(x + 20)}^{1} = x^{2}$

{(x + 20)}^{1} = x^{2}

${(x + 20)}^{1} = x^{2}$

Step 2.2.1.2

Simplify.

x + 20 = x^{2}

$x + 20 = x^{2}$

x + 20 = x^{2}

$x + 20 = x^{2}$

x + 20 = x^{2}

$x + 20 = x^{2}$

x + 20 = x^{2}

$x + 20 = x^{2}$

Step 3

Solve for

x

$x$ .

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Step 3.1

Subtract

x^{2}

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

x + 20 - x^{2} = 0

$x + 20 - x^{2} = 0$

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

Factor

- 1

$- 1$ out of

x + 20 - x^{2}

$x + 20 - x^{2}$ .

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Step 3.2.1.1

Reorder the expression.

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Step 3.2.1.1.1

Move

20

$20$ .

x - x^{2} + 20 = 0

$x - x^{2} + 20 = 0$

Step 3.2.1.1.2

Reorder

x

$x$ and

- x^{2}

$- x^{2}$ .

- x^{2} + x + 20 = 0

$- x^{2} + x + 20 = 0$

- x^{2} + x + 20 = 0

$- x^{2} + x + 20 = 0$

Step 3.2.1.2

Factor

- 1

$- 1$ out of

- x^{2}

$- x^{2}$ .

- (x^{2}) + x + 20 = 0

$- (x^{2}) + x + 20 = 0$

Step 3.2.1.3

Factor

- 1

$- 1$ out of

x

$x$ .

- (x^{2}) - 1 (- x) + 20 = 0

$- (x^{2}) - 1 (- x) + 20 = 0$

Step 3.2.1.4

Rewrite

20

$20$ as

- 1 (- 20)

$- 1 (- 20)$ .

- (x^{2}) - 1 (- x) - 1 \cdot - 20 = 0

$- (x^{2}) - 1 (- x) - 1 \cdot - 20 = 0$

Step 3.2.1.5

Factor

- 1

$- 1$ out of

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

$- (x^{2}) - 1 (- x)$ .

- (x^{2} - x) - 1 \cdot - 20 = 0

$- (x^{2} - x) - 1 \cdot - 20 = 0$

Step 3.2.1.6

Factor

- 1

$- 1$ out of

- (x^{2} - x) - 1 (- 20)

$- (x^{2} - x) - 1 (- 20)$ .

- (x^{2} - x - 20) = 0

$- (x^{2} - x - 20) = 0$

- (x^{2} - x - 20) = 0

$- (x^{2} - x - 20) = 0$

Step 3.2.2

Factor.

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Step 3.2.2.1

Factor

x^{2} - x - 20

$x^{2} - x - 20$ using the AC method.

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Step 3.2.2.1.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

- 20

$- 20$ and whose sum is

- 1

$- 1$ .

- 5, 4

$- 5, 4$

Step 3.2.2.1.2

Write the factored form using these integers.

- ((x - 5) (x + 4)) = 0

$- ((x - 5) (x + 4)) = 0$

- ((x - 5) (x + 4)) = 0

$- ((x - 5) (x + 4)) = 0$

Step 3.2.2.2

Remove unnecessary parentheses.

- (x - 5) (x + 4) = 0

$- (x - 5) (x + 4) = 0$

- (x - 5) (x + 4) = 0

$- (x - 5) (x + 4) = 0$

- (x - 5) (x + 4) = 0

$- (x - 5) (x + 4) = 0$

Step 3.3

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x - 5 = 0

$x - 5 = 0$

x + 4 = 0

$x + 4 = 0$

Step 3.4

Set

x - 5

$x - 5$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.4.1

Set

x - 5

$x - 5$ equal to

0

$0$ .

x - 5 = 0

$x - 5 = 0$

Step 3.4.2

Add

5

$5$ to both sides of the equation.

x = 5

$x = 5$

x = 5

$x = 5$

Step 3.5

Set

x + 4

$x + 4$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.5.1

Set

x + 4

$x + 4$ equal to

0

$0$ .

x + 4 = 0

$x + 4 = 0$

Step 3.5.2

Subtract

4

$4$ from both sides of the equation.

x = - 4

$x = - 4$

x = - 4

$x = - 4$

Step 3.6

The final solution is all the values that make

- (x - 5) (x + 4) = 0

$- (x - 5) (x + 4) = 0$ true.

x = 5, - 4

$x = 5, - 4$

x = 5, - 4

$x = 5, - 4$

Step 4

Exclude the solutions that do not make

\sqrt{x + 20} = x

$\sqrt{x + 20} = x$ true.

x = 5

$x = 5$