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

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

Solve for x

$$x=\frac{\sqrt{6}}{3}+2\approx 2.816496581$$

$$x=-\frac{\sqrt{6}}{3}+2\approx 1.183503419$$

Steps Using the Quadratic Formula

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$x^{2}-1=\left(2x-3\right)^{2}$$

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

$$x^{2}-1=4x^{2}-12x+9$$

Subtract 4x^{2} from both sides.

$$x^{2}-1-4x^{2}=-12x+9$$

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

$$-3x^{2}-1=-12x+9$$

Add 12x to both sides.

$$-3x^{2}-1+12x=9$$

Subtract 9 from both sides.

$$-3x^{2}-1+12x-9=0$$

Subtract 9 from -1 to get -10.

$$-3x^{2}-10+12x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-3x^{2}+12x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 12 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-3\right)\left(-10\right)}}{2\left(-3\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-3\right)\left(-10\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-12±\sqrt{144+12\left(-10\right)}}{2\left(-3\right)}$$

Multiply 12 times -10.

$$x=\frac{-12±\sqrt{144-120}}{2\left(-3\right)}$$

Add 144 to -120.

$$x=\frac{-12±\sqrt{24}}{2\left(-3\right)}$$

Take the square root of 24.

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

Multiply 2 times -3.

$$x=\frac{-12±2\sqrt{6}}{-6}$$

Now solve the equation x=\frac{-12±2\sqrt{6}}{-6} when ± is plus. Add -12 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}-12}{-6}$$

Divide -12+2\sqrt{6} by -6.

$$x=-\frac{\sqrt{6}}{3}+2$$

Now solve the equation x=\frac{-12±2\sqrt{6}}{-6} when ± is minus. Subtract 2\sqrt{6} from -12.

$$x=\frac{-2\sqrt{6}-12}{-6}$$

Divide -12-2\sqrt{6} by -6.

$$x=\frac{\sqrt{6}}{3}+2$$

The equation is now solved.

$$x=-\frac{\sqrt{6}}{3}+2 x=\frac{\sqrt{6}}{3}+2$$

Steps for Completing the Square

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$x^{2}-1=\left(2x-3\right)^{2}$$

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

$$x^{2}-1=4x^{2}-12x+9$$

Subtract 4x^{2} from both sides.

$$x^{2}-1-4x^{2}=-12x+9$$

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

$$-3x^{2}-1=-12x+9$$

Add 12x to both sides.

$$-3x^{2}-1+12x=9$$

Add 1 to both sides.

$$-3x^{2}+12x=9+1$$

Add 9 and 1 to get 10.

$$-3x^{2}+12x=10$$

Divide both sides by -3.

$$\frac{-3x^{2}+12x}{-3}=\frac{10}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{12}{-3}x=\frac{10}{-3}$$

Divide 12 by -3.

$$x^{2}-4x=\frac{10}{-3}$$

Divide 10 by -3.

$$x^{2}-4x=-\frac{10}{3}$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=-\frac{10}{3}+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-\frac{10}{3}+4$$

Add -\frac{10}{3} to 4.

$$x^{2}-4x+4=\frac{2}{3}$$

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{\frac{2}{3}}$$

Simplify.

$$x-2=\frac{\sqrt{6}}{3} x-2=-\frac{\sqrt{6}}{3}$$

Add 2 to both sides of the equation.

$$x=\frac{\sqrt{6}}{3}+2 x=-\frac{\sqrt{6}}{3}+2$$