Solve (2sqrt{2})^2=(sqrt{3}x)^2+x^2-2sqrt{3}x*x*frac{sqrt{3}}{3}

$$( 2 \sqrt { 2 } ) ^ { 2 } = ( \sqrt { 3 } x ) ^ { 2 } + x ^ { 2 } - 2 \sqrt { 3 } x \cdot x \cdot \frac { \sqrt { 3 } } { 3 }$$

Solve for x

$$x=2$$

$$x=-2$$

Steps Using Difference of Squares

Multiply both sides of the equation by 3.

$$3\times \left(2\sqrt{2}\right)^{2}=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\sqrt{3}xx\sqrt{3}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$3\times \left(2\sqrt{2}\right)^{2}=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3xx$$

Multiply x and x to get x^{2}.

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

Expand \left(2\sqrt{2}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$3\times 4\left(\sqrt{2}\right)^{2}=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3x^{2}$$

The square of \sqrt{2} is 2.

$$3\times 4\times 2=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3x^{2}$$

Multiply 4 and 2 to get 8.

$$3\times 8=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3x^{2}$$

Multiply 3 and 8 to get 24.

$$24=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3x^{2}$$

Expand \left(\sqrt{3}x\right)^{2}.

$$24=3\left(\left(\sqrt{3}\right)^{2}x^{2}+x^{2}\right)-2\times 3x^{2}$$

The square of \sqrt{3} is 3.

$$24=3\left(3x^{2}+x^{2}\right)-2\times 3x^{2}$$

Combine 3x^{2} and x^{2} to get 4x^{2}.

$$24=3\times 4x^{2}-2\times 3x^{2}$$

Multiply 3 and 4 to get 12.

$$24=12x^{2}-2\times 3x^{2}$$

Multiply 2 and 3 to get 6.

$$24=12x^{2}-6x^{2}$$

Combine 12x^{2} and -6x^{2} to get 6x^{2}.

$$24=6x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$6x^{2}=24$$

Subtract 24 from both sides.

$$6x^{2}-24=0$$

Divide both sides by 6.

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

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 of the equation by 3.

$$3\times \left(2\sqrt{2}\right)^{2}=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\sqrt{3}xx\sqrt{3}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$3\times \left(2\sqrt{2}\right)^{2}=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3xx$$

Multiply x and x to get x^{2}.

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

Expand \left(2\sqrt{2}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$3\times 4\left(\sqrt{2}\right)^{2}=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3x^{2}$$

The square of \sqrt{2} is 2.

$$3\times 4\times 2=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3x^{2}$$

Multiply 4 and 2 to get 8.

$$3\times 8=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3x^{2}$$

Multiply 3 and 8 to get 24.

$$24=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3x^{2}$$

Expand \left(\sqrt{3}x\right)^{2}.

$$24=3\left(\left(\sqrt{3}\right)^{2}x^{2}+x^{2}\right)-2\times 3x^{2}$$

The square of \sqrt{3} is 3.

$$24=3\left(3x^{2}+x^{2}\right)-2\times 3x^{2}$$

Combine 3x^{2} and x^{2} to get 4x^{2}.

$$24=3\times 4x^{2}-2\times 3x^{2}$$

Multiply 3 and 4 to get 12.

$$24=12x^{2}-2\times 3x^{2}$$

Multiply 2 and 3 to get 6.

$$24=12x^{2}-6x^{2}$$

Combine 12x^{2} and -6x^{2} to get 6x^{2}.

$$24=6x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$6x^{2}=24$$

Divide both sides by 6.

$$x^{2}=\frac{24}{6}$$

Divide 24 by 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 of the equation by 3.

$$3\times \left(2\sqrt{2}\right)^{2}=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\sqrt{3}xx\sqrt{3}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$3\times \left(2\sqrt{2}\right)^{2}=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3xx$$

Multiply x and x to get x^{2}.

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

Expand \left(2\sqrt{2}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$3\times 4\left(\sqrt{2}\right)^{2}=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3x^{2}$$

The square of \sqrt{2} is 2.

$$3\times 4\times 2=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3x^{2}$$

Multiply 4 and 2 to get 8.

$$3\times 8=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3x^{2}$$

Multiply 3 and 8 to get 24.

$$24=3\left(\left(\sqrt{3}x\right)^{2}+x^{2}\right)-2\times 3x^{2}$$

Expand \left(\sqrt{3}x\right)^{2}.

$$24=3\left(\left(\sqrt{3}\right)^{2}x^{2}+x^{2}\right)-2\times 3x^{2}$$

The square of \sqrt{3} is 3.

$$24=3\left(3x^{2}+x^{2}\right)-2\times 3x^{2}$$

Combine 3x^{2} and x^{2} to get 4x^{2}.

$$24=3\times 4x^{2}-2\times 3x^{2}$$

Multiply 3 and 4 to get 12.

$$24=12x^{2}-2\times 3x^{2}$$

Multiply 2 and 3 to get 6.

$$24=12x^{2}-6x^{2}$$

Combine 12x^{2} and -6x^{2} to get 6x^{2}.

$$24=6x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$6x^{2}=24$$

Subtract 24 from both sides.

$$6x^{2}-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 0 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 6\left(-24\right)}}{2\times 6}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 6\left(-24\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{0±\sqrt{-24\left(-24\right)}}{2\times 6}$$

Multiply -24 times -24.

$$x=\frac{0±\sqrt{576}}{2\times 6}$$

Take the square root of 576.

$$x=\frac{0±24}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{0±24}{12}$$

Now solve the equation x=\frac{0±24}{12} when ± is plus. Divide 24 by 12.

$$x=2$$

Now solve the equation x=\frac{0±24}{12} when ± is minus. Divide -24 by 12.

$$x=-2$$

The equation is now solved.

$$x=2 x=-2$$