Solve left(2x-1right)^2-left(x+1right)^2=9

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

Solve for x

$$x=-1$$

$$x=3$$

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}.

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

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

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

To find the opposite of x^{2}+2x+1, find the opposite of each term.

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-4x+1-2x-1=9$$

Combine -4x and -2x to get -6x.

$$3x^{2}-6x+1-1=9$$

Subtract 1 from 1 to get 0.

$$3x^{2}-6x=9$$

Subtract 9 from both sides.

$$3x^{2}-6x-9=0$$

Divide both sides by 3.

$$x^{2}-2x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-3\right)=-3$$

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. The only such pair is the system solution.

$$a=-3 b=1$$

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

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

Factor out x in x^{2}-3x.

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

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

$$\left(x-3\right)\left(x+1\right)$$

To find equation solutions, solve x-3=0 and x+1=0.

$$x=3 x=-1$$

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}.

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

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

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

To find the opposite of x^{2}+2x+1, find the opposite of each term.

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-4x+1-2x-1=9$$

Combine -4x and -2x to get -6x.

$$3x^{2}-6x+1-1=9$$

Subtract 1 from 1 to get 0.

$$3x^{2}-6x=9$$

Subtract 9 from both sides.

$$3x^{2}-6x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3\left(-9\right)}}{2\times 3}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 3\left(-9\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-6\right)±\sqrt{36-12\left(-9\right)}}{2\times 3}$$

Multiply -12 times -9.

$$x=\frac{-\left(-6\right)±\sqrt{36+108}}{2\times 3}$$

Add 36 to 108.

$$x=\frac{-\left(-6\right)±\sqrt{144}}{2\times 3}$$

Take the square root of 144.

$$x=\frac{-\left(-6\right)±12}{2\times 3}$$

The opposite of -6 is 6.

$$x=\frac{6±12}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{6±12}{6}$$

Now solve the equation x=\frac{6±12}{6} when ± is plus. Add 6 to 12.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{6±12}{6} when ± is minus. Subtract 12 from 6.

$$x=-\frac{6}{6}$$

Divide -6 by 6.

$$x=-1$$

The equation is now solved.

$$x=3 x=-1$$

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}.

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

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

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

To find the opposite of x^{2}+2x+1, find the opposite of each term.

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-4x+1-2x-1=9$$

Combine -4x and -2x to get -6x.

$$3x^{2}-6x+1-1=9$$

Subtract 1 from 1 to get 0.

$$3x^{2}-6x=9$$

Divide both sides by 3.

$$\frac{3x^{2}-6x}{3}=\frac{9}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{6}{3}\right)x=\frac{9}{3}$$

Divide -6 by 3.

$$x^{2}-2x=\frac{9}{3}$$

Divide 9 by 3.

$$x^{2}-2x=3$$

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

$$x^{2}-2x+1=3+1$$

Add 3 to 1.

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

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=2 x-1=-2$$

Add 1 to both sides of the equation.

$$x=3 x=-1$$