Solve left(6(x-1)right)^2=36x

$${ \left(6(x-1) \right) }^{ 2 } =36x$$

Solve for x

$$x = \frac{\sqrt{5} + 3}{2} \approx 2.618033989$$

$$x=\frac{3-\sqrt{5}}{2}\approx 0.381966011$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 6 by x-1.

$$\left(6x-6\right)^{2}=36x$$

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

$$36x^{2}-72x+36=36x$$

Subtract 36x from both sides.

$$36x^{2}-72x+36-36x=0$$

Combine -72x and -36x to get -108x.

$$36x^{2}-108x+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, -108 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-108\right)±\sqrt{\left(-108\right)^{2}-4\times 36\times 36}}{2\times 36}$$

Square -108.

$$x=\frac{-\left(-108\right)±\sqrt{11664-4\times 36\times 36}}{2\times 36}$$

Multiply -4 times 36.

$$x=\frac{-\left(-108\right)±\sqrt{11664-144\times 36}}{2\times 36}$$

Multiply -144 times 36.

$$x=\frac{-\left(-108\right)±\sqrt{11664-5184}}{2\times 36}$$

Add 11664 to -5184.

$$x=\frac{-\left(-108\right)±\sqrt{6480}}{2\times 36}$$

Take the square root of 6480.

$$x=\frac{-\left(-108\right)±36\sqrt{5}}{2\times 36}$$

The opposite of -108 is 108.

$$x=\frac{108±36\sqrt{5}}{2\times 36}$$

Multiply 2 times 36.

$$x=\frac{108±36\sqrt{5}}{72}$$

Now solve the equation x=\frac{108±36\sqrt{5}}{72} when ± is plus. Add 108 to 36\sqrt{5}.

$$x=\frac{36\sqrt{5}+108}{72}$$

Divide 108+36\sqrt{5} by 72.

$$x=\frac{\sqrt{5}+3}{2}$$

Now solve the equation x=\frac{108±36\sqrt{5}}{72} when ± is minus. Subtract 36\sqrt{5} from 108.

$$x=\frac{108-36\sqrt{5}}{72}$$

Divide 108-36\sqrt{5} by 72.

$$x=\frac{3-\sqrt{5}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{5}+3}{2} x=\frac{3-\sqrt{5}}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 6 by x-1.

$$\left(6x-6\right)^{2}=36x$$

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

$$36x^{2}-72x+36=36x$$

Subtract 36x from both sides.

$$36x^{2}-72x+36-36x=0$$

Combine -72x and -36x to get -108x.

$$36x^{2}-108x+36=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$36x^{2}-108x=-36$$

Divide both sides by 36.

$$\frac{36x^{2}-108x}{36}=-\frac{36}{36}$$

Dividing by 36 undoes the multiplication by 36.

$$x^{2}+\left(-\frac{108}{36}\right)x=-\frac{36}{36}$$

Divide -108 by 36.

$$x^{2}-3x=-\frac{36}{36}$$

Divide -36 by 36.

$$x^{2}-3x=-1$$

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-1+\frac{9}{4}$$

Add -1 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{5}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{5}}{2} x-\frac{3}{2}=-\frac{\sqrt{5}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{5}+3}{2} x=\frac{3-\sqrt{5}}{2}$$