Solve (2x)^2+(6-2sqrt{3}x)^2+6/sqrt{3}=12/sqrt{3}-2x

$$( 2 x ) ^ { 2 } + ( 6 - 2 \sqrt { 3 } x ) ^ { 2 } + \frac { 6 } { \sqrt { 3 } } = \frac { 12 } { \sqrt { 3 } } - 2 x$$

Solve for x (complex solution)

$$x=\frac{i\sqrt{143-8\sqrt{3}}}{16}+\frac{3\sqrt{3}}{4}-\frac{1}{16}\approx 1.236538106+0.710258518i$$

$$x=-\frac{i\sqrt{143-8\sqrt{3}}}{16}+\frac{3\sqrt{3}}{4}-\frac{1}{16}\approx 1.236538106-0.710258518i$$

Steps Using the Quadratic Formula

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

$$2^{2}x^{2}+\left(6-2\sqrt{3}x\right)^{2}+\frac{6}{\sqrt{3}}=\frac{12}{\sqrt{3}}-2x$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+\frac{6}{\sqrt{3}}=\frac{12}{\sqrt{3}}-2x$$

Rationalize the denominator of \frac{6}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+\frac{6\sqrt{3}}{\left(\sqrt{3}\right)^{2}}=\frac{12}{\sqrt{3}}-2x$$

The square of \sqrt{3} is 3.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+\frac{6\sqrt{3}}{3}=\frac{12}{\sqrt{3}}-2x$$

Divide 6\sqrt{3} by 3 to get 2\sqrt{3}.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+2\sqrt{3}=\frac{12}{\sqrt{3}}-2x$$

Rationalize the denominator of \frac{12}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+2\sqrt{3}=\frac{12\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-2x$$

The square of \sqrt{3} is 3.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+2\sqrt{3}=\frac{12\sqrt{3}}{3}-2x$$

Divide 12\sqrt{3} by 3 to get 4\sqrt{3}.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+2\sqrt{3}=4\sqrt{3}-2x$$

Subtract 4\sqrt{3} from both sides.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+2\sqrt{3}-4\sqrt{3}=-2x$$

Combine 2\sqrt{3} and -4\sqrt{3} to get -2\sqrt{3}.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}-2\sqrt{3}=-2x$$

Add 2x to both sides.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}-2\sqrt{3}+2x=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-2\sqrt{3}x\right)^{2}.

$$4x^{2}+36-24\sqrt{3}x+4\left(\sqrt{3}\right)^{2}x^{2}-2\sqrt{3}+2x=0$$

The square of \sqrt{3} is 3.

$$4x^{2}+36-24\sqrt{3}x+4\times 3x^{2}-2\sqrt{3}+2x=0$$

Multiply 4 and 3 to get 12.

$$4x^{2}+36-24\sqrt{3}x+12x^{2}-2\sqrt{3}+2x=0$$

Combine 4x^{2} and 12x^{2} to get 16x^{2}.

$$16x^{2}+36-24\sqrt{3}x-2\sqrt{3}+2x=0$$

Combine all terms containing x.

$$16x^{2}+36+\left(-24\sqrt{3}+2\right)x-2\sqrt{3}=0$$

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.

$$16x^{2}+\left(2-24\sqrt{3}\right)x+36-2\sqrt{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -24\sqrt{3}+2 for b, and 36-2\sqrt{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(2-24\sqrt{3}\right)±\sqrt{\left(2-24\sqrt{3}\right)^{2}-4\times 16\left(36-2\sqrt{3}\right)}}{2\times 16}$$

Square -24\sqrt{3}+2.

$$x=\frac{-\left(2-24\sqrt{3}\right)±\sqrt{1732-96\sqrt{3}-4\times 16\left(36-2\sqrt{3}\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(2-24\sqrt{3}\right)±\sqrt{1732-96\sqrt{3}-64\left(36-2\sqrt{3}\right)}}{2\times 16}$$

Multiply -64 times 36-2\sqrt{3}.

$$x=\frac{-\left(2-24\sqrt{3}\right)±\sqrt{1732-96\sqrt{3}+128\sqrt{3}-2304}}{2\times 16}$$

Add 1732-96\sqrt{3} to -2304+128\sqrt{3}.

$$x=\frac{-\left(2-24\sqrt{3}\right)±\sqrt{32\sqrt{3}-572}}{2\times 16}$$

Take the square root of -572+32\sqrt{3}.

$$x=\frac{-\left(2-24\sqrt{3}\right)±2i\sqrt{143-8\sqrt{3}}}{2\times 16}$$

The opposite of -24\sqrt{3}+2 is 24\sqrt{3}-2.

$$x=\frac{24\sqrt{3}-2±2i\sqrt{143-8\sqrt{3}}}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{24\sqrt{3}-2±2i\sqrt{143-8\sqrt{3}}}{32}$$

Now solve the equation x=\frac{24\sqrt{3}-2±2i\sqrt{143-8\sqrt{3}}}{32} when ± is plus. Add 24\sqrt{3}-2 to 2i\sqrt{143-8\sqrt{3}}.

$$x=\frac{24\sqrt{3}-2+2i\sqrt{143-8\sqrt{3}}}{32}$$

Divide 24\sqrt{3}-2+2i\sqrt{143-8\sqrt{3}} by 32.

$$x=\frac{i\sqrt{143-8\sqrt{3}}}{16}+\frac{3\sqrt{3}}{4}-\frac{1}{16}$$

Now solve the equation x=\frac{24\sqrt{3}-2±2i\sqrt{143-8\sqrt{3}}}{32} when ± is minus. Subtract 2i\sqrt{143-8\sqrt{3}} from 24\sqrt{3}-2.

$$x=\frac{-2i\sqrt{143-8\sqrt{3}}+24\sqrt{3}-2}{32}$$

Divide 24\sqrt{3}-2-2i\sqrt{143-8\sqrt{3}} by 32.

$$x=-\frac{i\sqrt{143-8\sqrt{3}}}{16}+\frac{3\sqrt{3}}{4}-\frac{1}{16}$$

The equation is now solved.

$$x=\frac{i\sqrt{143-8\sqrt{3}}}{16}+\frac{3\sqrt{3}}{4}-\frac{1}{16} x=-\frac{i\sqrt{143-8\sqrt{3}}}{16}+\frac{3\sqrt{3}}{4}-\frac{1}{16}$$

Steps for Completing the Square

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

$$2^{2}x^{2}+\left(6-2\sqrt{3}x\right)^{2}+\frac{6}{\sqrt{3}}=\frac{12}{\sqrt{3}}-2x$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+\frac{6}{\sqrt{3}}=\frac{12}{\sqrt{3}}-2x$$

Rationalize the denominator of \frac{6}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+\frac{6\sqrt{3}}{\left(\sqrt{3}\right)^{2}}=\frac{12}{\sqrt{3}}-2x$$

The square of \sqrt{3} is 3.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+\frac{6\sqrt{3}}{3}=\frac{12}{\sqrt{3}}-2x$$

Divide 6\sqrt{3} by 3 to get 2\sqrt{3}.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+2\sqrt{3}=\frac{12}{\sqrt{3}}-2x$$

Rationalize the denominator of \frac{12}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+2\sqrt{3}=\frac{12\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-2x$$

The square of \sqrt{3} is 3.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+2\sqrt{3}=\frac{12\sqrt{3}}{3}-2x$$

Divide 12\sqrt{3} by 3 to get 4\sqrt{3}.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+2\sqrt{3}=4\sqrt{3}-2x$$

Add 2x to both sides.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+2\sqrt{3}+2x=4\sqrt{3}$$

Subtract 2\sqrt{3} from both sides.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+2x=4\sqrt{3}-2\sqrt{3}$$

Combine 4\sqrt{3} and -2\sqrt{3} to get 2\sqrt{3}.

$$4x^{2}+\left(6-2\sqrt{3}x\right)^{2}+2x=2\sqrt{3}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-2\sqrt{3}x\right)^{2}.

$$4x^{2}+36-24\sqrt{3}x+4\left(\sqrt{3}\right)^{2}x^{2}+2x=2\sqrt{3}$$

The square of \sqrt{3} is 3.

$$4x^{2}+36-24\sqrt{3}x+4\times 3x^{2}+2x=2\sqrt{3}$$

Multiply 4 and 3 to get 12.

$$4x^{2}+36-24\sqrt{3}x+12x^{2}+2x=2\sqrt{3}$$

Combine 4x^{2} and 12x^{2} to get 16x^{2}.

$$16x^{2}+36-24\sqrt{3}x+2x=2\sqrt{3}$$

Subtract 36 from both sides.

$$16x^{2}-24\sqrt{3}x+2x=2\sqrt{3}-36$$

Combine all terms containing x.

$$16x^{2}+\left(-24\sqrt{3}+2\right)x=2\sqrt{3}-36$$

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.

$$16x^{2}+\left(2-24\sqrt{3}\right)x=2\sqrt{3}-36$$

Divide both sides by 16.

$$\frac{16x^{2}+\left(2-24\sqrt{3}\right)x}{16}=\frac{2\sqrt{3}-36}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\frac{2-24\sqrt{3}}{16}x=\frac{2\sqrt{3}-36}{16}$$

Divide -24\sqrt{3}+2 by 16.

$$x^{2}+\left(-\frac{3\sqrt{3}}{2}+\frac{1}{8}\right)x=\frac{2\sqrt{3}-36}{16}$$

Divide 2\sqrt{3}-36 by 16.

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

Divide -\frac{3\sqrt{3}}{2}+\frac{1}{8}, the coefficient of the x term, by 2 to get -\frac{3\sqrt{3}}{4}+\frac{1}{16}. Then add the square of -\frac{3\sqrt{3}}{4}+\frac{1}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\left(-\frac{3\sqrt{3}}{2}+\frac{1}{8}\right)x+\left(-\frac{3\sqrt{3}}{4}+\frac{1}{16}\right)^{2}=\frac{\sqrt{3}}{8}-\frac{9}{4}+\left(-\frac{3\sqrt{3}}{4}+\frac{1}{16}\right)^{2}$$

Square -\frac{3\sqrt{3}}{4}+\frac{1}{16}.

$$x^{2}+\left(-\frac{3\sqrt{3}}{2}+\frac{1}{8}\right)x-\frac{3\sqrt{3}}{32}+\frac{433}{256}=\frac{\sqrt{3}}{8}-\frac{9}{4}-\frac{3\sqrt{3}}{32}+\frac{433}{256}$$

Add \frac{\sqrt{3}}{8}-\frac{9}{4} to \frac{433}{256}-\frac{3\sqrt{3}}{32}.

$$x^{2}+\left(-\frac{3\sqrt{3}}{2}+\frac{1}{8}\right)x-\frac{3\sqrt{3}}{32}+\frac{433}{256}=\frac{\sqrt{3}}{32}-\frac{143}{256}$$

Factor x^{2}+\left(-\frac{3\sqrt{3}}{2}+\frac{1}{8}\right)x-\frac{3\sqrt{3}}{32}+\frac{433}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3\sqrt{3}}{4}+\frac{1}{16}\right)^{2}=\frac{\sqrt{3}}{32}-\frac{143}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3\sqrt{3}}{4}+\frac{1}{16}\right)^{2}}=\sqrt{\frac{\sqrt{3}}{32}-\frac{143}{256}}$$

Simplify.

$$x-\frac{3\sqrt{3}}{4}+\frac{1}{16}=\frac{i\sqrt{143-8\sqrt{3}}}{16} x-\frac{3\sqrt{3}}{4}+\frac{1}{16}=-\frac{i\sqrt{143-8\sqrt{3}}}{16}$$

Subtract -\frac{3\sqrt{3}}{4}+\frac{1}{16} from both sides of the equation.

$$x=\frac{i\sqrt{143-8\sqrt{3}}}{16}+\frac{3\sqrt{3}}{4}-\frac{1}{16} x=-\frac{i\sqrt{143-8\sqrt{3}}}{16}+\frac{3\sqrt{3}}{4}-\frac{1}{16}$$