Solve 16=(2x-1)^2

$$16 = ( 2 x - 1 ) ^ { 2 }$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps Using Factoring By Grouping

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

$$16=4x^{2}-4x+1$$

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}-4x+1=16$$

Subtract 16 from both sides.

$$4x^{2}-4x+1-16=0$$

Subtract 16 from 1 to get -15.

$$4x^{2}-4x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=-4 ab=4\left(-15\right)=-60$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -4.

$$a=-10 b=6$$

Rewrite 4x^{2}-4x-15 as \left(4x^{2}-10x\right)+\left(6x-15\right).

$$\left(4x^{2}-10x\right)+\left(6x-15\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(2x-5\right)+3\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(2x+3\right)$$

To find equation solutions, solve 2x-5=0 and 2x+3=0.

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

Steps Using the Quadratic Formula

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

$$16=4x^{2}-4x+1$$

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}-4x+1=16$$

Subtract 16 from both sides.

$$4x^{2}-4x+1-16=0$$

Subtract 16 from 1 to get -15.

$$4x^{2}-4x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -4 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4\left(-15\right)}}{2\times 4}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 4\left(-15\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16-16\left(-15\right)}}{2\times 4}$$

Multiply -16 times -15.

$$x=\frac{-\left(-4\right)±\sqrt{16+240}}{2\times 4}$$

Add 16 to 240.

$$x=\frac{-\left(-4\right)±\sqrt{256}}{2\times 4}$$

Take the square root of 256.

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

The opposite of -4 is 4.

$$x=\frac{4±16}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{4±16}{8}$$

Now solve the equation x=\frac{4±16}{8} when ± is plus. Add 4 to 16.

$$x=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{4±16}{8} when ± is minus. Subtract 16 from 4.

$$x=-\frac{12}{8}$$

Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$16=4x^{2}-4x+1$$

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}-4x+1=16$$

Subtract 1 from both sides.

$$4x^{2}-4x=16-1$$

Subtract 1 from 16 to get 15.

$$4x^{2}-4x=15$$

Divide both sides by 4.

$$\frac{4x^{2}-4x}{4}=\frac{15}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{4}{4}\right)x=\frac{15}{4}$$

Divide -4 by 4.

$$x^{2}-x=\frac{15}{4}$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{15}{4}+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=\frac{15+1}{4}$$

Add \frac{15}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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