Precalculus Examples

$2 \sqrt{x + 4} = x + 1$

Step 1

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

${(2 \sqrt{x + 4})}^{2} = {(x + 1)}^{2}$

Step 2

Simplify each side of the equation.

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Step 2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 4}$ as ${(x + 4)}^{\frac{1}{2}}$ .

${(2 {(x + 4)}^{\frac{1}{2}})}^{2} = {(x + 1)}^{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify ${(2 {(x + 4)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.1.1

Apply the product rule to $2 {(x + 4)}^{\frac{1}{2}}$ .

$2^{2} {({(x + 4)}^{\frac{1}{2}})}^{2} = {(x + 1)}^{2}$

Step 2.2.1.2

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

$4 {({(x + 4)}^{\frac{1}{2}})}^{2} = {(x + 1)}^{2}$

Step 2.2.1.3

Multiply the exponents in ${({(x + 4)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.1.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 {(x + 4)}^{\frac{1}{2} \cdot 2} = {(x + 1)}^{2}$

Step 2.2.1.3.2

Cancel the common factor of $2$ .

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Step 2.2.1.3.2.1

Cancel the common factor.

$4 {(x + 4)}^{\frac{1}{2} \cdot 2} = {(x + 1)}^{2}$

Step 2.2.1.3.2.2

Rewrite the expression.

$4 {(x + 4)}^{1} = {(x + 1)}^{2}$

Step 2.2.1.4

Simplify.

$4 (x + 4) = {(x + 1)}^{2}$

Step 2.2.1.5

Apply the distributive property.

$4 x + 4 \cdot 4 = {(x + 1)}^{2}$

Step 2.2.1.6

Multiply $4$ by $4$ .

$4 x + 16 = {(x + 1)}^{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify ${(x + 1)}^{2}$ .

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Step 2.3.1.1

Rewrite ${(x + 1)}^{2}$ as $(x + 1) (x + 1)$ .

$4 x + 16 = (x + 1) (x + 1)$

Step 2.3.1.2

Expand $(x + 1) (x + 1)$ using the FOIL Method.

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Step 2.3.1.2.1

Apply the distributive property.

$4 x + 16 = x (x + 1) + 1 (x + 1)$

Step 2.3.1.2.2

Apply the distributive property.

$4 x + 16 = x \cdot x + x \cdot 1 + 1 (x + 1)$

Step 2.3.1.2.3

Apply the distributive property.

$4 x + 16 = x \cdot x + x \cdot 1 + 1 x + 1 \cdot 1$

Step 2.3.1.3

Simplify and combine like terms.

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Step 2.3.1.3.1

Simplify each term.

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Step 2.3.1.3.1.1

Multiply $x$ by $x$ .

$4 x + 16 = x^{2} + x \cdot 1 + 1 x + 1 \cdot 1$

Step 2.3.1.3.1.2

Multiply $x$ by $1$ .

$4 x + 16 = x^{2} + x + 1 x + 1 \cdot 1$

Step 2.3.1.3.1.3

Multiply $x$ by $1$ .

$4 x + 16 = x^{2} + x + x + 1 \cdot 1$

Step 2.3.1.3.1.4

Multiply $1$ by $1$ .

$4 x + 16 = x^{2} + x + x + 1$

Step 2.3.1.3.2

Add $x$ and $x$ .

$4 x + 16 = x^{2} + 2 x + 1$

Step 3

Solve for $x$ .

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Step 3.1

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

$x^{2} + 2 x + 1 = 4 x + 16$

Step 3.2

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

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Step 3.2.1

Subtract $4 x$ from both sides of the equation.

$x^{2} + 2 x + 1 - 4 x = 16$

Step 3.2.2

Subtract $4 x$ from $2 x$ .

$x^{2} - 2 x + 1 = 16$

Step 3.3

Subtract $16$ from both sides of the equation.

$x^{2} - 2 x + 1 - 16 = 0$

Step 3.4

Subtract $16$ from $1$ .

$x^{2} - 2 x - 15 = 0$

Step 3.5

Factor $x^{2} - 2 x - 15$ using the AC method.

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Step 3.5.1

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 $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Step 3.5.2

Write the factored form using these integers.

$(x - 5) (x + 3) = 0$

Step 3.6

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

$x - 5 = 0$

$x + 3 = 0$

Step 3.7

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

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Step 3.7.1

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

$x - 5 = 0$

Step 3.7.2

Add $5$ to both sides of the equation.

$x = 5$

Step 3.8

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 3.8.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 3.8.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 3.9

The final solution is all the values that make $(x - 5) (x + 3) = 0$ true.

$x = 5, - 3$

Step 4

Exclude the solutions that do not make $2 \sqrt{x + 4} = x + 1$ true.

$x = 5$