Solve 1/3(x+2)^2-1/2-4-x^2/6=3/2(x+frac{1}{3})^2

$$\frac { 1 } { 3 } ( x + 2 ) ^ { 2 } - \frac { 1 } { 2 } - \frac { 4 - x ^ { 2 } } { 6 } = \frac { 3 } { 2 } ( x + \frac { 1 } { 3 } ) ^ { 2 }$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=0$$

Steps Using Factoring

Multiply both sides of the equation by 6, the least common multiple of 3,2,6.

$$2\left(x+2\right)^{2}-3-\left(4-x^{2}\right)=9\left(x+\frac{1}{3}\right)^{2}$$

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

$$2\left(x^{2}+4x+4\right)-3-\left(4-x^{2}\right)=9\left(x+\frac{1}{3}\right)^{2}$$

Use the distributive property to multiply 2 by x^{2}+4x+4.

$$2x^{2}+8x+8-3-\left(4-x^{2}\right)=9\left(x+\frac{1}{3}\right)^{2}$$

Subtract 3 from 8 to get 5.

$$2x^{2}+8x+5-\left(4-x^{2}\right)=9\left(x+\frac{1}{3}\right)^{2}$$

To find the opposite of 4-x^{2}, find the opposite of each term.

$$2x^{2}+8x+5-4+x^{2}=9\left(x+\frac{1}{3}\right)^{2}$$

Subtract 4 from 5 to get 1.

$$2x^{2}+8x+1+x^{2}=9\left(x+\frac{1}{3}\right)^{2}$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+8x+1=9\left(x+\frac{1}{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{1}{3}\right)^{2}.

$$3x^{2}+8x+1=9\left(x^{2}+\frac{2}{3}x+\frac{1}{9}\right)$$

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

$$3x^{2}+8x+1=9x^{2}+6x+1$$

Subtract 9x^{2} from both sides.

$$3x^{2}+8x+1-9x^{2}=6x+1$$

Combine 3x^{2} and -9x^{2} to get -6x^{2}.

$$-6x^{2}+8x+1=6x+1$$

Subtract 6x from both sides.

$$-6x^{2}+8x+1-6x=1$$

Combine 8x and -6x to get 2x.

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

Subtract 1 from both sides.

$$-6x^{2}+2x+1-1=0$$

Subtract 1 from 1 to get 0.

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

Factor out x.

$$x\left(-6x+2\right)=0$$

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

$$x=0 x=\frac{1}{3}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6, the least common multiple of 3,2,6.

$$2\left(x+2\right)^{2}-3-\left(4-x^{2}\right)=9\left(x+\frac{1}{3}\right)^{2}$$

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

$$2\left(x^{2}+4x+4\right)-3-\left(4-x^{2}\right)=9\left(x+\frac{1}{3}\right)^{2}$$

Use the distributive property to multiply 2 by x^{2}+4x+4.

$$2x^{2}+8x+8-3-\left(4-x^{2}\right)=9\left(x+\frac{1}{3}\right)^{2}$$

Subtract 3 from 8 to get 5.

$$2x^{2}+8x+5-\left(4-x^{2}\right)=9\left(x+\frac{1}{3}\right)^{2}$$

To find the opposite of 4-x^{2}, find the opposite of each term.

$$2x^{2}+8x+5-4+x^{2}=9\left(x+\frac{1}{3}\right)^{2}$$

Subtract 4 from 5 to get 1.

$$2x^{2}+8x+1+x^{2}=9\left(x+\frac{1}{3}\right)^{2}$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+8x+1=9\left(x+\frac{1}{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{1}{3}\right)^{2}.

$$3x^{2}+8x+1=9\left(x^{2}+\frac{2}{3}x+\frac{1}{9}\right)$$

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

$$3x^{2}+8x+1=9x^{2}+6x+1$$

Subtract 9x^{2} from both sides.

$$3x^{2}+8x+1-9x^{2}=6x+1$$

Combine 3x^{2} and -9x^{2} to get -6x^{2}.

$$-6x^{2}+8x+1=6x+1$$

Subtract 6x from both sides.

$$-6x^{2}+8x+1-6x=1$$

Combine 8x and -6x to get 2x.

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

Subtract 1 from both sides.

$$-6x^{2}+2x+1-1=0$$

Subtract 1 from 1 to get 0.

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

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

$$x=\frac{-2±\sqrt{2^{2}}}{2\left(-6\right)}$$

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-2±2}{-12}$$

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

$$x=\frac{0}{-12}$$

Divide 0 by -12.

$$x=0$$

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

$$x=-\frac{4}{-12}$$

Reduce the fraction \frac{-4}{-12} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{3}$$

The equation is now solved.

$$x=0 x=\frac{1}{3}$$

Steps for Completing the Square

Multiply both sides of the equation by 6, the least common multiple of 3,2,6.

$$2\left(x+2\right)^{2}-3-\left(4-x^{2}\right)=9\left(x+\frac{1}{3}\right)^{2}$$

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

$$2\left(x^{2}+4x+4\right)-3-\left(4-x^{2}\right)=9\left(x+\frac{1}{3}\right)^{2}$$

Use the distributive property to multiply 2 by x^{2}+4x+4.

$$2x^{2}+8x+8-3-\left(4-x^{2}\right)=9\left(x+\frac{1}{3}\right)^{2}$$

Subtract 3 from 8 to get 5.

$$2x^{2}+8x+5-\left(4-x^{2}\right)=9\left(x+\frac{1}{3}\right)^{2}$$

To find the opposite of 4-x^{2}, find the opposite of each term.

$$2x^{2}+8x+5-4+x^{2}=9\left(x+\frac{1}{3}\right)^{2}$$

Subtract 4 from 5 to get 1.

$$2x^{2}+8x+1+x^{2}=9\left(x+\frac{1}{3}\right)^{2}$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+8x+1=9\left(x+\frac{1}{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{1}{3}\right)^{2}.

$$3x^{2}+8x+1=9\left(x^{2}+\frac{2}{3}x+\frac{1}{9}\right)$$

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

$$3x^{2}+8x+1=9x^{2}+6x+1$$

Subtract 9x^{2} from both sides.

$$3x^{2}+8x+1-9x^{2}=6x+1$$

Combine 3x^{2} and -9x^{2} to get -6x^{2}.

$$-6x^{2}+8x+1=6x+1$$

Subtract 6x from both sides.

$$-6x^{2}+8x+1-6x=1$$

Combine 8x and -6x to get 2x.

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

Divide both sides by -6.

$$\frac{-6x^{2}+2x}{-6}=\frac{0}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{2}{-6}x=\frac{0}{-6}$$

Reduce the fraction \frac{2}{-6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{3}x=\frac{0}{-6}$$

Divide 0 by -6.

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

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{1}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{1}{6} x-\frac{1}{6}=-\frac{1}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{1}{3} x=0$$