Solve x^2+(3+3)x+8=0

$${ x }^{ 2 } +(3+3)x+8=0$$

Solve for x

$$x=-4$$

$$x=-2$$

Steps Using Factoring

Add 3 and 3 to get 6.

$$x^{2}+6x+8=0$$

To solve the equation, factor x^{2}+6x+8 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.

$$a+b=6 ab=8$$

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 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$a=2 b=4$$

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

$$\left(x+2\right)\left(x+4\right)$$

To find equation solutions, solve x+2=0 and x+4=0.

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

Steps Using Factoring By Grouping

Add 3 and 3 to get 6.

$$x^{2}+6x+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\times 8=8$$

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 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$a=2 b=4$$

Rewrite x^{2}+6x+8 as \left(x^{2}+2x\right)+\left(4x+8\right).

$$\left(x^{2}+2x\right)+\left(4x+8\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x+2\right)+4\left(x+2\right)$$

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

$$\left(x+2\right)\left(x+4\right)$$

To find equation solutions, solve x+2=0 and x+4=0.

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

Steps Using the Quadratic Formula

Add 3 and 3 to get 6.

$$x^{2}+6x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 8}}{2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 8}}{2}$$

Multiply -4 times 8.

$$x=\frac{-6±\sqrt{36-32}}{2}$$

Add 36 to -32.

$$x=\frac{-6±\sqrt{4}}{2}$$

Take the square root of 4.

$$x=\frac{-6±2}{2}$$

Now solve the equation x=\frac{-6±2}{2} when ± is plus. Add -6 to 2.

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

Divide -4 by 2.

$$x=-2$$

Now solve the equation x=\frac{-6±2}{2} when ± is minus. Subtract 2 from -6.

$$x=-\frac{8}{2}$$

Divide -8 by 2.

$$x=-4$$

The equation is now solved.

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

Steps for Completing the Square

Add 3 and 3 to get 6.

$$x^{2}+6x+8=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+6x=-8$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=-8+3^{2}$$

Square 3.

$$x^{2}+6x+9=-8+9$$

Add -8 to 9.

$$x^{2}+6x+9=1$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{1}$$

Simplify.

$$x+3=1 x+3=-1$$

Subtract 3 from both sides of the equation.

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