Solve 5(3+x)=1/3(-3-x)^2

$$5 ( 3 + x ) = \frac { 1 } { 3 } ( - 3 - x ) ^ { 2 }$$

Solve for x

$$x=-3$$

$$x=12$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5 by 3+x.

$$15+5x=\frac{1}{3}\left(-3-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-3-x\right)^{2}.

$$15+5x=\frac{1}{3}\left(9+6x+x^{2}\right)$$

Use the distributive property to multiply \frac{1}{3} by 9+6x+x^{2}.

$$15+5x=3+2x+\frac{1}{3}x^{2}$$

Subtract 3 from both sides.

$$15+5x-3=2x+\frac{1}{3}x^{2}$$

Subtract 3 from 15 to get 12.

$$12+5x=2x+\frac{1}{3}x^{2}$$

Subtract 2x from both sides.

$$12+5x-2x=\frac{1}{3}x^{2}$$

Combine 5x and -2x to get 3x.

$$12+3x=\frac{1}{3}x^{2}$$

Subtract \frac{1}{3}x^{2} from both sides.

$$12+3x-\frac{1}{3}x^{2}=0$$

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.

$$-\frac{1}{3}x^{2}+3x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{3} for a, 3 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-\frac{1}{3}\right)\times 12}}{2\left(-\frac{1}{3}\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-\frac{1}{3}\right)\times 12}}{2\left(-\frac{1}{3}\right)}$$

Multiply -4 times -\frac{1}{3}.

$$x=\frac{-3±\sqrt{9+\frac{4}{3}\times 12}}{2\left(-\frac{1}{3}\right)}$$

Multiply \frac{4}{3} times 12.

$$x=\frac{-3±\sqrt{9+16}}{2\left(-\frac{1}{3}\right)}$$

Add 9 to 16.

$$x=\frac{-3±\sqrt{25}}{2\left(-\frac{1}{3}\right)}$$

Take the square root of 25.

$$x=\frac{-3±5}{2\left(-\frac{1}{3}\right)}$$

Multiply 2 times -\frac{1}{3}.

$$x=\frac{-3±5}{-\frac{2}{3}}$$

Now solve the equation x=\frac{-3±5}{-\frac{2}{3}} when ± is plus. Add -3 to 5.

$$x=\frac{2}{-\frac{2}{3}}$$

Divide 2 by -\frac{2}{3} by multiplying 2 by the reciprocal of -\frac{2}{3}.

$$x=-3$$

Now solve the equation x=\frac{-3±5}{-\frac{2}{3}} when ± is minus. Subtract 5 from -3.

$$x=-\frac{8}{-\frac{2}{3}}$$

Divide -8 by -\frac{2}{3} by multiplying -8 by the reciprocal of -\frac{2}{3}.

$$x=12$$

The equation is now solved.

$$x=-3 x=12$$

Steps for Completing the Square

Use the distributive property to multiply 5 by 3+x.

$$15+5x=\frac{1}{3}\left(-3-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-3-x\right)^{2}.

$$15+5x=\frac{1}{3}\left(9+6x+x^{2}\right)$$

Use the distributive property to multiply \frac{1}{3} by 9+6x+x^{2}.

$$15+5x=3+2x+\frac{1}{3}x^{2}$$

Subtract 2x from both sides.

$$15+5x-2x=3+\frac{1}{3}x^{2}$$

Combine 5x and -2x to get 3x.

$$15+3x=3+\frac{1}{3}x^{2}$$

Subtract \frac{1}{3}x^{2} from both sides.

$$15+3x-\frac{1}{3}x^{2}=3$$

Subtract 15 from both sides.

$$3x-\frac{1}{3}x^{2}=3-15$$

Subtract 15 from 3 to get -12.

$$3x-\frac{1}{3}x^{2}=-12$$

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.

$$-\frac{1}{3}x^{2}+3x=-12$$

Multiply both sides by -3.

$$\frac{-\frac{1}{3}x^{2}+3x}{-\frac{1}{3}}=-\frac{12}{-\frac{1}{3}}$$

Dividing by -\frac{1}{3} undoes the multiplication by -\frac{1}{3}.

$$x^{2}+\frac{3}{-\frac{1}{3}}x=-\frac{12}{-\frac{1}{3}}$$

Divide 3 by -\frac{1}{3} by multiplying 3 by the reciprocal of -\frac{1}{3}.

$$x^{2}-9x=-\frac{12}{-\frac{1}{3}}$$

Divide -12 by -\frac{1}{3} by multiplying -12 by the reciprocal of -\frac{1}{3}.

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

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=36+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=36+\frac{81}{4}$$

Add 36 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{225}{4}$$

Factor x^{2}-9x+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{15}{2} x-\frac{9}{2}=-\frac{15}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=12 x=-3$$