Solve (3x-1)^2-(2x+1)^2=7

$$( 3 x - 1 ) ^ { 2 } - ( 2 x + 1 ) ^ { 2 } = 7$$

Solve for x

$$x=\frac{2\sqrt{15}}{5}+1\approx 2.549193338$$

$$x=-\frac{2\sqrt{15}}{5}+1\approx -0.549193338$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$9x^{2}-6x+1-\left(2x+1\right)^{2}=7$$

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

$$9x^{2}-6x+1-\left(4x^{2}+4x+1\right)=7$$

To find the opposite of 4x^{2}+4x+1, find the opposite of each term.

$$9x^{2}-6x+1-4x^{2}-4x-1=7$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}-6x+1-4x-1=7$$

Combine -6x and -4x to get -10x.

$$5x^{2}-10x+1-1=7$$

Subtract 1 from 1 to get 0.

$$5x^{2}-10x=7$$

Subtract 7 from both sides.

$$5x^{2}-10x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -10 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 5\left(-7\right)}}{2\times 5}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 5\left(-7\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-10\right)±\sqrt{100-20\left(-7\right)}}{2\times 5}$$

Multiply -20 times -7.

$$x=\frac{-\left(-10\right)±\sqrt{100+140}}{2\times 5}$$

Add 100 to 140.

$$x=\frac{-\left(-10\right)±\sqrt{240}}{2\times 5}$$

Take the square root of 240.

$$x=\frac{-\left(-10\right)±4\sqrt{15}}{2\times 5}$$

The opposite of -10 is 10.

$$x=\frac{10±4\sqrt{15}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{10±4\sqrt{15}}{10}$$

Now solve the equation x=\frac{10±4\sqrt{15}}{10} when ± is plus. Add 10 to 4\sqrt{15}.

$$x=\frac{4\sqrt{15}+10}{10}$$

Divide 10+4\sqrt{15} by 10.

$$x=\frac{2\sqrt{15}}{5}+1$$

Now solve the equation x=\frac{10±4\sqrt{15}}{10} when ± is minus. Subtract 4\sqrt{15} from 10.

$$x=\frac{10-4\sqrt{15}}{10}$$

Divide 10-4\sqrt{15} by 10.

$$x=-\frac{2\sqrt{15}}{5}+1$$

The equation is now solved.

$$x=\frac{2\sqrt{15}}{5}+1 x=-\frac{2\sqrt{15}}{5}+1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$9x^{2}-6x+1-\left(2x+1\right)^{2}=7$$

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

$$9x^{2}-6x+1-\left(4x^{2}+4x+1\right)=7$$

To find the opposite of 4x^{2}+4x+1, find the opposite of each term.

$$9x^{2}-6x+1-4x^{2}-4x-1=7$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}-6x+1-4x-1=7$$

Combine -6x and -4x to get -10x.

$$5x^{2}-10x+1-1=7$$

Subtract 1 from 1 to get 0.

$$5x^{2}-10x=7$$

Divide both sides by 5.

$$\frac{5x^{2}-10x}{5}=\frac{7}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{10}{5}\right)x=\frac{7}{5}$$

Divide -10 by 5.

$$x^{2}-2x=\frac{7}{5}$$

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.

$$x^{2}-2x+1=\frac{7}{5}+1$$

Add \frac{7}{5} to 1.

$$x^{2}-2x+1=\frac{12}{5}$$

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}.

$$\left(x-1\right)^{2}=\frac{12}{5}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{\frac{12}{5}}$$

Simplify.

$$x-1=\frac{2\sqrt{15}}{5} x-1=-\frac{2\sqrt{15}}{5}$$

Add 1 to both sides of the equation.

$$x=\frac{2\sqrt{15}}{5}+1 x=-\frac{2\sqrt{15}}{5}+1$$