Solve (2x+1)^2-9=0

$$( 2 x + 1 ) ^ { 2 } - 9 = 0$$

Solve for x

$$x=1$$

$$x=-2$$

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}.

$$4x^{2}+4x+1-9=0$$

Subtract 9 from 1 to get -8.

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

Divide both sides by 4.

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

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

$$a+b=1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=2$$

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

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

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

$$x\left(x-1\right)+2\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+2\right)$$

To find equation solutions, solve x-1=0 and x+2=0.

$$x=1 x=-2$$

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}.

$$4x^{2}+4x+1-9=0$$

Subtract 9 from 1 to get -8.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 4 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 4\left(-8\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-4±\sqrt{16-16\left(-8\right)}}{2\times 4}$$

Multiply -16 times -8.

$$x=\frac{-4±\sqrt{16+128}}{2\times 4}$$

Add 16 to 128.

$$x=\frac{-4±\sqrt{144}}{2\times 4}$$

Take the square root of 144.

$$x=\frac{-4±12}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-4±12}{8}$$

Now solve the equation x=\frac{-4±12}{8} when ± is plus. Add -4 to 12.

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

Divide 8 by 8.

$$x=1$$

Now solve the equation x=\frac{-4±12}{8} when ± is minus. Subtract 12 from -4.

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

Divide -16 by 8.

$$x=-2$$

The equation is now solved.

$$x=1 x=-2$$

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}.

$$4x^{2}+4x+1-9=0$$

Subtract 9 from 1 to get -8.

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

Add 8 to both sides. Anything plus zero gives itself.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide 4 by 4.

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

Divide 8 by 4.

$$x^{2}+x=2$$

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.

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

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

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

Add 2 to \frac{1}{4}.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=1 x=-2$$