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

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

Solve for x

$$x=1$$

Steps Using the Quadratic Formula

To raise \frac{\sqrt{3}}{6} to a power, raise both numerator and denominator to the power and then divide.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}+4\times \frac{\left(\sqrt{3}\right)^{2}}{6^{2}}\left(x^{2}-8x+16\right)=4$$

Express 4\times \frac{\left(\sqrt{3}\right)^{2}}{6^{2}} as a single fraction.

$$x^{2}+\frac{4\left(\sqrt{3}\right)^{2}}{6^{2}}\left(x^{2}-8x+16\right)=4$$

Express \frac{4\left(\sqrt{3}\right)^{2}}{6^{2}}\left(x^{2}-8x+16\right) as a single fraction.

$$x^{2}+\frac{4\left(\sqrt{3}\right)^{2}\left(x^{2}-8x+16\right)}{6^{2}}=4$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2} times \frac{6^{2}}{6^{2}}.

$$\frac{x^{2}\times 6^{2}}{6^{2}}+\frac{4\left(\sqrt{3}\right)^{2}\left(x^{2}-8x+16\right)}{6^{2}}=4$$

Since \frac{x^{2}\times 6^{2}}{6^{2}} and \frac{4\left(\sqrt{3}\right)^{2}\left(x^{2}-8x+16\right)}{6^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}\times 6^{2}+4\left(\sqrt{3}\right)^{2}\left(x^{2}-8x+16\right)}{6^{2}}=4$$

Calculate 6 to the power of 2 and get 36.

$$\frac{x^{2}\times 36+4\left(\sqrt{3}\right)^{2}\left(x^{2}-8x+16\right)}{6^{2}}=4$$

The square of \sqrt{3} is 3.

$$\frac{x^{2}\times 36+4\times 3\left(x^{2}-8x+16\right)}{6^{2}}=4$$

Multiply 4 and 3 to get 12.

$$\frac{x^{2}\times 36+12\left(x^{2}-8x+16\right)}{6^{2}}=4$$

Calculate 6 to the power of 2 and get 36.

$$\frac{x^{2}\times 36+12\left(x^{2}-8x+16\right)}{36}=4$$

Divide each term of x^{2}\times 36+12\left(x^{2}-8x+16\right) by 36 to get x^{2}+\frac{1}{3}\left(x^{2}-8x+16\right).

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

Use the distributive property to multiply \frac{1}{3} by x^{2}-8x+16.

$$x^{2}+\frac{1}{3}x^{2}-\frac{8}{3}x+\frac{16}{3}=4$$

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

$$\frac{4}{3}x^{2}-\frac{8}{3}x+\frac{16}{3}=4$$

Subtract 4 from both sides.

$$\frac{4}{3}x^{2}-\frac{8}{3}x+\frac{16}{3}-4=0$$

Subtract 4 from \frac{16}{3} to get \frac{4}{3}.

$$\frac{4}{3}x^{2}-\frac{8}{3}x+\frac{4}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{4}{3} for a, -\frac{8}{3} for b, and \frac{4}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -\frac{8}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{64}{9}-4\times \frac{4}{3}\times \frac{4}{3}}}{2\times \frac{4}{3}}$$

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

$$x=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{64}{9}-\frac{16}{3}\times \frac{4}{3}}}{2\times \frac{4}{3}}$$

Multiply -\frac{16}{3} times \frac{4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{64-64}{9}}}{2\times \frac{4}{3}}$$

Add \frac{64}{9} to -\frac{64}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{8}{3}\right)±\sqrt{0}}{2\times \frac{4}{3}}$$

Take the square root of 0.

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

The opposite of -\frac{8}{3} is \frac{8}{3}.

$$x=\frac{\frac{8}{3}}{2\times \frac{4}{3}}$$

Multiply 2 times \frac{4}{3}.

$$x=\frac{\frac{8}{3}}{\frac{8}{3}}$$

Divide \frac{8}{3} by \frac{8}{3} by multiplying \frac{8}{3} by the reciprocal of \frac{8}{3}.

$$x=1$$

Steps for Completing the Square

To raise \frac{\sqrt{3}}{6} to a power, raise both numerator and denominator to the power and then divide.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}+4\times \frac{\left(\sqrt{3}\right)^{2}}{6^{2}}\left(x^{2}-8x+16\right)=4$$

Express 4\times \frac{\left(\sqrt{3}\right)^{2}}{6^{2}} as a single fraction.

$$x^{2}+\frac{4\left(\sqrt{3}\right)^{2}}{6^{2}}\left(x^{2}-8x+16\right)=4$$

Express \frac{4\left(\sqrt{3}\right)^{2}}{6^{2}}\left(x^{2}-8x+16\right) as a single fraction.

$$x^{2}+\frac{4\left(\sqrt{3}\right)^{2}\left(x^{2}-8x+16\right)}{6^{2}}=4$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2} times \frac{6^{2}}{6^{2}}.

$$\frac{x^{2}\times 6^{2}}{6^{2}}+\frac{4\left(\sqrt{3}\right)^{2}\left(x^{2}-8x+16\right)}{6^{2}}=4$$

Since \frac{x^{2}\times 6^{2}}{6^{2}} and \frac{4\left(\sqrt{3}\right)^{2}\left(x^{2}-8x+16\right)}{6^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}\times 6^{2}+4\left(\sqrt{3}\right)^{2}\left(x^{2}-8x+16\right)}{6^{2}}=4$$

Calculate 6 to the power of 2 and get 36.

$$\frac{x^{2}\times 36+4\left(\sqrt{3}\right)^{2}\left(x^{2}-8x+16\right)}{6^{2}}=4$$

The square of \sqrt{3} is 3.

$$\frac{x^{2}\times 36+4\times 3\left(x^{2}-8x+16\right)}{6^{2}}=4$$

Multiply 4 and 3 to get 12.

$$\frac{x^{2}\times 36+12\left(x^{2}-8x+16\right)}{6^{2}}=4$$

Calculate 6 to the power of 2 and get 36.

$$\frac{x^{2}\times 36+12\left(x^{2}-8x+16\right)}{36}=4$$

Divide each term of x^{2}\times 36+12\left(x^{2}-8x+16\right) by 36 to get x^{2}+\frac{1}{3}\left(x^{2}-8x+16\right).

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

Use the distributive property to multiply \frac{1}{3} by x^{2}-8x+16.

$$x^{2}+\frac{1}{3}x^{2}-\frac{8}{3}x+\frac{16}{3}=4$$

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

$$\frac{4}{3}x^{2}-\frac{8}{3}x+\frac{16}{3}=4$$

Subtract \frac{16}{3} from both sides.

$$\frac{4}{3}x^{2}-\frac{8}{3}x=4-\frac{16}{3}$$

Subtract \frac{16}{3} from 4 to get -\frac{4}{3}.

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

Divide both sides of the equation by \frac{4}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by \frac{4}{3} undoes the multiplication by \frac{4}{3}.

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

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

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

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

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

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

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

Add -1 to 1.

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

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

$$\left(x-1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-1=0 x-1=0$$

Add 1 to both sides of the equation.

$$x=1 x=1$$

The equation is now solved. Solutions are the same.

$$x=1$$