Solve (3x-1)^2=x(4x-1)

$$( 3 x - 1 ) ^ { 2 } = x ( 4 x - 1 )$$

Solve for x

$$x=\frac{\sqrt{5}}{10}+\frac{1}{2}\approx 0.723606798$$

$$x=-\frac{\sqrt{5}}{10}+\frac{1}{2}\approx 0.276393202$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$9x^{2}-6x+1=x\left(4x-1\right)$$

Use the distributive property to multiply x by 4x-1.

$$9x^{2}-6x+1=4x^{2}-x$$

Subtract 4x^{2} from both sides.

$$9x^{2}-6x+1-4x^{2}=-x$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}-6x+1=-x$$

Add x to both sides.

$$5x^{2}-6x+1+x=0$$

Combine -6x and x to get -5x.

$$5x^{2}-5x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -5 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 5}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-5\right)±\sqrt{25-20}}{2\times 5}$$

Add 25 to -20.

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

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{5}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{5±\sqrt{5}}{10}$$

Now solve the equation x=\frac{5±\sqrt{5}}{10} when ± is plus. Add 5 to \sqrt{5}.

$$x=\frac{\sqrt{5}+5}{10}$$

Divide 5+\sqrt{5} by 10.

$$x=\frac{\sqrt{5}}{10}+\frac{1}{2}$$

Now solve the equation x=\frac{5±\sqrt{5}}{10} when ± is minus. Subtract \sqrt{5} from 5.

$$x=\frac{5-\sqrt{5}}{10}$$

Divide 5-\sqrt{5} by 10.

$$x=-\frac{\sqrt{5}}{10}+\frac{1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{5}}{10}+\frac{1}{2} x=-\frac{\sqrt{5}}{10}+\frac{1}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$9x^{2}-6x+1=x\left(4x-1\right)$$

Use the distributive property to multiply x by 4x-1.

$$9x^{2}-6x+1=4x^{2}-x$$

Subtract 4x^{2} from both sides.

$$9x^{2}-6x+1-4x^{2}=-x$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}-6x+1=-x$$

Add x to both sides.

$$5x^{2}-6x+1+x=0$$

Combine -6x and x to get -5x.

$$5x^{2}-5x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$5x^{2}-5x=-1$$

Divide both sides by 5.

$$\frac{5x^{2}-5x}{5}=-\frac{1}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{5}{5}\right)x=-\frac{1}{5}$$

Divide -5 by 5.

$$x^{2}-x=-\frac{1}{5}$$

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{1}{5}+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=-\frac{1}{5}+\frac{1}{4}$$

Add -\frac{1}{5} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{1}{20}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{1}{20}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{20}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{5}}{10} x-\frac{1}{2}=-\frac{\sqrt{5}}{10}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{5}}{10}+\frac{1}{2} x=-\frac{\sqrt{5}}{10}+\frac{1}{2}$$