Solve (2)4(2x-1)^2=36

$$( 2 ) 4 ( 2 x - 1 ) ^ { 2 } = 36$$

Solve for x

$$x=\frac{3\sqrt{2}}{4}+\frac{1}{2}\approx 1.560660172$$

$$x=-\frac{3\sqrt{2}}{4}+\frac{1}{2}\approx -0.560660172$$

Steps Using the Quadratic Formula

Multiply 2 and 4 to get 8.

$$8\left(2x-1\right)^{2}=36$$

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

$$8\left(4x^{2}-4x+1\right)=36$$

Use the distributive property to multiply 8 by 4x^{2}-4x+1.

$$32x^{2}-32x+8=36$$

Subtract 36 from both sides.

$$32x^{2}-32x+8-36=0$$

Subtract 36 from 8 to get -28.

$$32x^{2}-32x-28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 32 for a, -32 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 32\left(-28\right)}}{2\times 32}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 32\left(-28\right)}}{2\times 32}$$

Multiply -4 times 32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-128\left(-28\right)}}{2\times 32}$$

Multiply -128 times -28.

$$x=\frac{-\left(-32\right)±\sqrt{1024+3584}}{2\times 32}$$

Add 1024 to 3584.

$$x=\frac{-\left(-32\right)±\sqrt{4608}}{2\times 32}$$

Take the square root of 4608.

$$x=\frac{-\left(-32\right)±48\sqrt{2}}{2\times 32}$$

The opposite of -32 is 32.

$$x=\frac{32±48\sqrt{2}}{2\times 32}$$

Multiply 2 times 32.

$$x=\frac{32±48\sqrt{2}}{64}$$

Now solve the equation x=\frac{32±48\sqrt{2}}{64} when ± is plus. Add 32 to 48\sqrt{2}.

$$x=\frac{48\sqrt{2}+32}{64}$$

Divide 32+48\sqrt{2} by 64.

$$x=\frac{3\sqrt{2}}{4}+\frac{1}{2}$$

Now solve the equation x=\frac{32±48\sqrt{2}}{64} when ± is minus. Subtract 48\sqrt{2} from 32.

$$x=\frac{32-48\sqrt{2}}{64}$$

Divide 32-48\sqrt{2} by 64.

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

The equation is now solved.

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

Steps for Completing the Square

Multiply 2 and 4 to get 8.

$$8\left(2x-1\right)^{2}=36$$

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

$$8\left(4x^{2}-4x+1\right)=36$$

Use the distributive property to multiply 8 by 4x^{2}-4x+1.

$$32x^{2}-32x+8=36$$

Subtract 8 from both sides.

$$32x^{2}-32x=36-8$$

Subtract 8 from 36 to get 28.

$$32x^{2}-32x=28$$

Divide both sides by 32.

$$\frac{32x^{2}-32x}{32}=\frac{28}{32}$$

Dividing by 32 undoes the multiplication by 32.

$$x^{2}+\left(-\frac{32}{32}\right)x=\frac{28}{32}$$

Divide -32 by 32.

$$x^{2}-x=\frac{28}{32}$$

Reduce the fraction \frac{28}{32} to lowest terms by extracting and canceling out 4.

$$x^{2}-x=\frac{7}{8}$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{7}{8}+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=\frac{7}{8}+\frac{1}{4}$$

Add \frac{7}{8} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{9}{8}$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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