Solve (2x-1)^2-9=16

$$( 2 x - 1 ) ^ { 2 } - 9 = 16$$

Solve for x

$$x=3$$

$$x=-2$$

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}.

$$4x^{2}-4x+1-9=16$$

Subtract 9 from 1 to get -8.

$$4x^{2}-4x-8=16$$

Subtract 16 from both sides.

$$4x^{2}-4x-8-16=0$$

Subtract 16 from -8 to get -24.

$$4x^{2}-4x-24=0$$

Divide both sides by 4.

$$x^{2}-x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-6\right)=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite x^{2}-x-6 as \left(x^{2}-3x\right)+\left(2x-6\right).

$$\left(x^{2}-3x\right)+\left(2x-6\right)$$

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

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

Factor out common term x-3 by using distributive property.

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

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

$$x=3 x=-2$$

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}.

$$4x^{2}-4x+1-9=16$$

Subtract 9 from 1 to get -8.

$$4x^{2}-4x-8=16$$

Subtract 16 from both sides.

$$4x^{2}-4x-8-16=0$$

Subtract 16 from -8 to get -24.

$$4x^{2}-4x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -4 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4\left(-24\right)}}{2\times 4}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 4\left(-24\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16-16\left(-24\right)}}{2\times 4}$$

Multiply -16 times -24.

$$x=\frac{-\left(-4\right)±\sqrt{16+384}}{2\times 4}$$

Add 16 to 384.

$$x=\frac{-\left(-4\right)±\sqrt{400}}{2\times 4}$$

Take the square root of 400.

$$x=\frac{-\left(-4\right)±20}{2\times 4}$$

The opposite of -4 is 4.

$$x=\frac{4±20}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{4±20}{8}$$

Now solve the equation x=\frac{4±20}{8} when ± is plus. Add 4 to 20.

$$x=\frac{24}{8}$$

Divide 24 by 8.

$$x=3$$

Now solve the equation x=\frac{4±20}{8} when ± is minus. Subtract 20 from 4.

$$x=-\frac{16}{8}$$

Divide -16 by 8.

$$x=-2$$

The equation is now solved.

$$x=3 x=-2$$

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}.

$$4x^{2}-4x+1-9=16$$

Subtract 9 from 1 to get -8.

$$4x^{2}-4x-8=16$$

Add 8 to both sides.

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

Add 16 and 8 to get 24.

$$4x^{2}-4x=24$$

Divide both sides by 4.

$$\frac{4x^{2}-4x}{4}=\frac{24}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{4}{4}\right)x=\frac{24}{4}$$

Divide -4 by 4.

$$x^{2}-x=\frac{24}{4}$$

Divide 24 by 4.

$$x^{2}-x=6$$

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.

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

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

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

Add 6 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{25}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

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

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

$$x=3 x=-2$$