How to solve (x+4)^(2)=4

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (x+4)^(2)=4 | Equatix

Question

$${ \left(x+4 \right) }^{ 2 } =4$$

Solve for x

$x=-2$
$x=-6$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+4\right)^{2}$.

$$x^{2}+8x+16=4$$

Subtract $4$ from both sides.

$$x^{2}+8x+16-4=0$$

Subtract $4$ from $16$ to get $12$.

$$x^{2}+8x+12=0$$

To solve the equation, factor $x^{2}+8x+12$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.

$$a+b=8$$ $$ab=12$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $12$.

$$1,12$$ $$2,6$$ $$3,4$$

Calculate the sum for each pair.

$$1+12=13$$ $$2+6=8$$ $$3+4=7$$

The solution is the pair that gives sum $8$.

$$a=2$$ $$b=6$$

Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.

$$\left(x+2\right)\left(x+6\right)$$

To find equation solutions, solve $x+2=0$ and $x+6=0$.

$$x=-2$$ $$x=-6$$

Steps Using Factoring By Grouping

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+4\right)^{2}$.

$$x^{2}+8x+16=4$$

Subtract $4$ from both sides.

$$x^{2}+8x+16-4=0$$

Subtract $4$ from $16$ to get $12$.

$$x^{2}+8x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+12$. To find $a$ and $b$, set up a system to be solved.

$$a+b=8$$ $$ab=1\times 12=12$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $12$.

$$1,12$$ $$2,6$$ $$3,4$$

Calculate the sum for each pair.

$$1+12=13$$ $$2+6=8$$ $$3+4=7$$

The solution is the pair that gives sum $8$.

$$a=2$$ $$b=6$$

Rewrite $x^{2}+8x+12$ as $\left(x^{2}+2x\right)+\left(6x+12\right)$.

$$\left(x^{2}+2x\right)+\left(6x+12\right)$$

Factor out $x$ in the first and $6$ in the second group.

$$x\left(x+2\right)+6\left(x+2\right)$$

Factor out common term $x+2$ by using distributive property.

$$\left(x+2\right)\left(x+6\right)$$

To find equation solutions, solve $x+2=0$ and $x+6=0$.

$$x=-2$$ $$x=-6$$

Steps Using the Quadratic Formula

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+4\right)^{2}$.

$$x^{2}+8x+16=4$$

Subtract $4$ from both sides.

$$x^{2}+8x+16-4=0$$

Subtract $4$ from $16$ to get $12$.

$$x^{2}+8x+12=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $8$ for $b$, and $12$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-8±\sqrt{8^{2}-4\times 12}}{2}$$

Square $8$.

$$x=\frac{-8±\sqrt{64-4\times 12}}{2}$$

Multiply $-4$ times $12$.

$$x=\frac{-8±\sqrt{64-48}}{2}$$

Add $64$ to $-48$.

$$x=\frac{-8±\sqrt{16}}{2}$$

Take the square root of $16$.

$$x=\frac{-8±4}{2}$$

Now solve the equation $x=\frac{-8±4}{2}$ when $±$ is plus. Add $-8$ to $4$.

$$x=-\frac{4}{2}$$

Divide $-4$ by $2$.

$$x=-2$$

Now solve the equation $x=\frac{-8±4}{2}$ when $±$ is minus. Subtract $4$ from $-8$.

$$x=-\frac{12}{2}$$

Divide $-12$ by $2$.

$$x=-6$$

The equation is now solved.

$$x=-2$$ $$x=-6$$

Steps for Completing the Square

Take the square root of both sides of the equation.

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

Simplify.

$$x+4=2$$ $$x+4=-2$$

Subtract $4$ from both sides of the equation.

$$x=-2$$ $$x=-6$$

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Example

Question

Solve for x

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square