Algebra Examples

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

Step 1

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

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

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

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

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

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

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

Step 2.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.2.1.1.2.1

Cancel the common factor.

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

Step 2.2.1.1.2.2

Rewrite the expression.

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

Step 2.2.1.2

Simplify.

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

Step 2.3

Simplify the right side.

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Step 2.3.1

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

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Step 2.3.1.1

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

$x + 2 = (2 x + 1) (2 x + 1)$

Step 2.3.1.2

Expand $(2 x + 1) (2 x + 1)$ using the FOIL Method.

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Step 2.3.1.2.1

Apply the distributive property.

$x + 2 = 2 x (2 x + 1) + 1 (2 x + 1)$

Step 2.3.1.2.2

Apply the distributive property.

$x + 2 = 2 x (2 x) + 2 x \cdot 1 + 1 (2 x + 1)$

Step 2.3.1.2.3

Apply the distributive property.

$x + 2 = 2 x (2 x) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1$

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

Rewrite using the commutative property of multiplication.

$x + 2 = 2 \cdot 2 x \cdot x + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1$

Step 2.3.1.3.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 2.3.1.3.1.2.1

Move $x$ .

$x + 2 = 2 \cdot 2 (x \cdot x) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1$

Step 2.3.1.3.1.2.2

Multiply $x$ by $x$ .

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

Step 2.3.1.3.1.3

Multiply $2$ by $2$ .

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

Step 2.3.1.3.1.4

Multiply $2$ by $1$ .

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

Step 2.3.1.3.1.5

Multiply $2 x$ by $1$ .

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

Step 2.3.1.3.1.6

Multiply $1$ by $1$ .

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

Step 2.3.1.3.2

Add $2 x$ and $2 x$ .

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

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.

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

Step 3.2

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

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Step 3.2.1

Subtract $x$ from both sides of the equation.

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

Step 3.2.2

Subtract $x$ from $4 x$ .

$4 x^{2} + 3 x + 1 = 2$

Step 3.3

Subtract $2$ from both sides of the equation.

$4 x^{2} + 3 x + 1 - 2 = 0$

Step 3.4

Subtract $2$ from $1$ .

$4 x^{2} + 3 x - 1 = 0$

Step 3.5

Factor by grouping.

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Step 3.5.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot - 1 = - 4$ and whose sum is $b = 3$ .

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Step 3.5.1.1

Factor $3$ out of $3 x$ .

$4 x^{2} + 3 (x) - 1 = 0$

Step 3.5.1.2

Rewrite $3$ as $- 1$ plus $4$

$4 x^{2} + (- 1 + 4) x - 1 = 0$

Step 3.5.1.3

Apply the distributive property.

$4 x^{2} - 1 x + 4 x - 1 = 0$

Step 3.5.2

Factor out the greatest common factor from each group.

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Step 3.5.2.1

Group the first two terms and the last two terms.

$(4 x^{2} - 1 x) + 4 x - 1 = 0$

Step 3.5.2.2

Factor out the greatest common factor (GCF) from each group.

$x (4 x - 1) + 1 (4 x - 1) = 0$

Step 3.5.3

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

$(4 x - 1) (x + 1) = 0$

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 $0$ .

$4 x - 1 = 0$

$x + 1 = 0$

Step 3.7

Set $4 x - 1$ equal to $0$ and solve for $x$ .

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Step 3.7.1

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Step 3.7.2

Solve $4 x - 1 = 0$ for $x$ .

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Step 3.7.2.1

Add $1$ to both sides of the equation.

$4 x = 1$

Step 3.7.2.2

Divide each term in $4 x = 1$ by $4$ and simplify.

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Step 3.7.2.2.1

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Step 3.7.2.2.2

Simplify the left side.

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Step 3.7.2.2.2.1

Cancel the common factor of $4$ .

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Step 3.7.2.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Step 3.7.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Step 3.8

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 3.8.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 3.8.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.9

The final solution is all the values that make $(4 x - 1) (x + 1) = 0$ true.

$x = \frac{1}{4}, - 1$

Step 4

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

$x = \frac{1}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{4}$

Decimal Form:

$x = 0.25$