Solve (6+x)^2/3=3

$$\frac { ( 6 + x ) ^ { 2 } } { 3 } = 3$$

Solve for x

$$x=-3$$

$$x=-9$$

Steps Using Factoring

Multiply both sides by 3.

$$\left(6+x\right)^{2}=3\times 3$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6+x\right)^{2}.

$$36+12x+x^{2}=3\times 3$$

Multiply 3 and 3 to get 9.

$$36+12x+x^{2}=9$$

Subtract 9 from both sides.

$$36+12x+x^{2}-9=0$$

Subtract 9 from 36 to get 27.

$$27+12x+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+12x+27=0$$

To solve the equation, factor x^{2}+12x+27 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.

$$a+b=12 ab=27$$

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 27.

$$1,27 3,9$$

Calculate the sum for each pair.

$$1+27=28 3+9=12$$

The solution is the pair that gives sum 12.

$$a=3 b=9$$

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

$$\left(x+3\right)\left(x+9\right)$$

To find equation solutions, solve x+3=0 and x+9=0.

$$x=-3 x=-9$$

Steps Using Factoring By Grouping

Multiply both sides by 3.

$$\left(6+x\right)^{2}=3\times 3$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6+x\right)^{2}.

$$36+12x+x^{2}=3\times 3$$

Multiply 3 and 3 to get 9.

$$36+12x+x^{2}=9$$

Subtract 9 from both sides.

$$36+12x+x^{2}-9=0$$

Subtract 9 from 36 to get 27.

$$27+12x+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+12x+27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+27. To find a and b, set up a system to be solved.

$$a+b=12 ab=1\times 27=27$$

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 27.

$$1,27 3,9$$

Calculate the sum for each pair.

$$1+27=28 3+9=12$$

The solution is the pair that gives sum 12.

$$a=3 b=9$$

Rewrite x^{2}+12x+27 as \left(x^{2}+3x\right)+\left(9x+27\right).

$$\left(x^{2}+3x\right)+\left(9x+27\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(x+3\right)+9\left(x+3\right)$$

Factor out common term x+3 by using distributive property.

$$\left(x+3\right)\left(x+9\right)$$

To find equation solutions, solve x+3=0 and x+9=0.

$$x=-3 x=-9$$

Steps Using the Quadratic Formula

Multiply both sides by 3.

$$\left(6+x\right)^{2}=3\times 3$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6+x\right)^{2}.

$$36+12x+x^{2}=3\times 3$$

Multiply 3 and 3 to get 9.

$$36+12x+x^{2}=9$$

Subtract 9 from both sides.

$$36+12x+x^{2}-9=0$$

Subtract 9 from 36 to get 27.

$$27+12x+x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+12x+27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 12 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 27}}{2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 27}}{2}$$

Multiply -4 times 27.

$$x=\frac{-12±\sqrt{144-108}}{2}$$

Add 144 to -108.

$$x=\frac{-12±\sqrt{36}}{2}$$

Take the square root of 36.

$$x=\frac{-12±6}{2}$$

Now solve the equation x=\frac{-12±6}{2} when ± is plus. Add -12 to 6.

$$x=-\frac{6}{2}$$

Divide -6 by 2.

$$x=-3$$

Now solve the equation x=\frac{-12±6}{2} when ± is minus. Subtract 6 from -12.

$$x=-\frac{18}{2}$$

Divide -18 by 2.

$$x=-9$$

The equation is now solved.

$$x=-3 x=-9$$

Steps for Completing the Square

Multiply both sides by 3.

$$\left(6+x\right)^{2}=3\times 3$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6+x\right)^{2}.

$$36+12x+x^{2}=3\times 3$$

Multiply 3 and 3 to get 9.

$$36+12x+x^{2}=9$$

Subtract 36 from both sides.

$$12x+x^{2}=9-36$$

Subtract 36 from 9 to get -27.

$$12x+x^{2}=-27$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}+12x=-27$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+12x+6^{2}=-27+6^{2}$$

Square 6.

$$x^{2}+12x+36=-27+36$$

Add -27 to 36.

$$x^{2}+12x+36=9$$

Factor x^{2}+12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+6\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{9}$$

Simplify.

$$x+6=3 x+6=-3$$

Subtract 6 from both sides of the equation.

$$x=-3 x=-9$$