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

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

Solve for x

$$x=4\sqrt{3}-6\approx 0.92820323$$

$$x=-4\sqrt{3}-6\approx -12.92820323$$

Steps Using the Quadratic Formula

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

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

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

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

The square of \sqrt{3} is 3.

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

To raise \frac{x\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2} times \frac{3^{2}}{3^{2}}.

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

Since \frac{x^{2}\times 3^{2}}{3^{2}} and \frac{\left(x\sqrt{3}\right)^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

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

Calculate 3 to the power of 2 and get 9.

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

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

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

The square of \sqrt{3} is 3.

$$4-4x+x^{2}=\frac{x^{2}\times 9+x^{2}\times 3}{3^{2}}$$

Combine x^{2}\times 9 and x^{2}\times 3 to get 12x^{2}.

$$4-4x+x^{2}=\frac{12x^{2}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$4-4x+x^{2}=\frac{12x^{2}}{9}$$

Divide 12x^{2} by 9 to get \frac{4}{3}x^{2}.

$$4-4x+x^{2}=\frac{4}{3}x^{2}$$

Subtract \frac{4}{3}x^{2} from both sides.

$$4-4x+x^{2}-\frac{4}{3}x^{2}=0$$

Combine x^{2} and -\frac{4}{3}x^{2} to get -\frac{1}{3}x^{2}.

$$4-4x-\frac{1}{3}x^{2}=0$$

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.

$$-\frac{1}{3}x^{2}-4x+4=0$$

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-\frac{1}{3}\right)\times 4}}{2\left(-\frac{1}{3}\right)}$$

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

$$x=\frac{-\left(-4\right)±\sqrt{16+\frac{4}{3}\times 4}}{2\left(-\frac{1}{3}\right)}$$

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

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

Add 16 to \frac{16}{3}.

$$x=\frac{-\left(-4\right)±\sqrt{\frac{64}{3}}}{2\left(-\frac{1}{3}\right)}$$

Take the square root of \frac{64}{3}.

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

The opposite of -4 is 4.

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

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

$$x=\frac{4±\frac{8\sqrt{3}}{3}}{-\frac{2}{3}}$$

Now solve the equation x=\frac{4±\frac{8\sqrt{3}}{3}}{-\frac{2}{3}} when ± is plus. Add 4 to \frac{8\sqrt{3}}{3}.

$$x=\frac{\frac{8\sqrt{3}}{3}+4}{-\frac{2}{3}}$$

Divide 4+\frac{8\sqrt{3}}{3} by -\frac{2}{3} by multiplying 4+\frac{8\sqrt{3}}{3} by the reciprocal of -\frac{2}{3}.

$$x=-4\sqrt{3}-6$$

Now solve the equation x=\frac{4±\frac{8\sqrt{3}}{3}}{-\frac{2}{3}} when ± is minus. Subtract \frac{8\sqrt{3}}{3} from 4.

$$x=\frac{-\frac{8\sqrt{3}}{3}+4}{-\frac{2}{3}}$$

Divide 4-\frac{8\sqrt{3}}{3} by -\frac{2}{3} by multiplying 4-\frac{8\sqrt{3}}{3} by the reciprocal of -\frac{2}{3}.

$$x=4\sqrt{3}-6$$

The equation is now solved.

$$x=-4\sqrt{3}-6 x=4\sqrt{3}-6$$

Steps for Completing the Square

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

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

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

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

The square of \sqrt{3} is 3.

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

To raise \frac{x\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2} times \frac{3^{2}}{3^{2}}.

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

Since \frac{x^{2}\times 3^{2}}{3^{2}} and \frac{\left(x\sqrt{3}\right)^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

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

Calculate 3 to the power of 2 and get 9.

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

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

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

The square of \sqrt{3} is 3.

$$4-4x+x^{2}=\frac{x^{2}\times 9+x^{2}\times 3}{3^{2}}$$

Combine x^{2}\times 9 and x^{2}\times 3 to get 12x^{2}.

$$4-4x+x^{2}=\frac{12x^{2}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$4-4x+x^{2}=\frac{12x^{2}}{9}$$

Divide 12x^{2} by 9 to get \frac{4}{3}x^{2}.

$$4-4x+x^{2}=\frac{4}{3}x^{2}$$

Subtract \frac{4}{3}x^{2} from both sides.

$$4-4x+x^{2}-\frac{4}{3}x^{2}=0$$

Combine x^{2} and -\frac{4}{3}x^{2} to get -\frac{1}{3}x^{2}.

$$4-4x-\frac{1}{3}x^{2}=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-4x-\frac{1}{3}x^{2}=-4$$

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.

$$-\frac{1}{3}x^{2}-4x=-4$$

Multiply both sides by -3.

$$\frac{-\frac{1}{3}x^{2}-4x}{-\frac{1}{3}}=-\frac{4}{-\frac{1}{3}}$$

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

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

Divide -4 by -\frac{1}{3} by multiplying -4 by the reciprocal of -\frac{1}{3}.

$$x^{2}+12x=-\frac{4}{-\frac{1}{3}}$$

Divide -4 by -\frac{1}{3} by multiplying -4 by the reciprocal of -\frac{1}{3}.

$$x^{2}+12x=12$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+12x+6^{2}=12+6^{2}$$

Square 6.

$$x^{2}+12x+36=12+36$$

Add 12 to 36.

$$x^{2}+12x+36=48$$

Factor x^{2}+12x+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}.

$$\left(x+6\right)^{2}=48$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{48}$$

Simplify.

$$x+6=4\sqrt{3} x+6=-4\sqrt{3}$$

Subtract 6 from both sides of the equation.

$$x=4\sqrt{3}-6 x=-4\sqrt{3}-6$$