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

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

Solve for x

$$x=2$$

$$x=-2$$

Steps Using Difference of Squares

Multiply both sides by 3, the reciprocal of \frac{1}{3}.

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

Multiply 2 and 3 to get 6.

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

Subtract 6 from both sides.

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

Subtract 6 from 2 to get -4.

$$x^{2}-4=0$$

Consider x^{2}-4. Rewrite x^{2}-4 as x^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

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

$$x=2 x=-2$$

Steps by Finding Square Root

Multiply both sides by 3, the reciprocal of \frac{1}{3}.

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

Multiply 2 and 3 to get 6.

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

Subtract 2 from both sides.

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

Subtract 2 from 6 to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

Multiply both sides by 3, the reciprocal of \frac{1}{3}.

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

Multiply 2 and 3 to get 6.

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

Subtract 6 from both sides.

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

Subtract 6 from 2 to get -4.

$$x^{2}-4=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-4\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{0±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{0±4}{2}$$

Now solve the equation x=\frac{0±4}{2} when ± is plus. Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{0±4}{2} when ± is minus. Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=2 x=-2$$