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

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

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=\frac{1}{4}=0.25$$

Steps Using Factoring By Grouping

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

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

Subtract x^{2} from both sides.

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

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}=-2x+1$$

Add 2x to both sides.

$$8x^{2}+2x=1$$

Subtract 1 from both sides.

$$8x^{2}+2x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=2 ab=8\left(-1\right)=-8$$

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. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=-2 b=4$$

Rewrite 8x^{2}+2x-1 as \left(8x^{2}-2x\right)+\left(4x-1\right).

$$\left(8x^{2}-2x\right)+\left(4x-1\right)$$

Factor out 2x in 8x^{2}-2x.

$$2x\left(4x-1\right)+4x-1$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(2x+1\right)$$

To find equation solutions, solve 4x-1=0 and 2x+1=0.

$$x=\frac{1}{4} x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

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

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

Subtract x^{2} from both sides.

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

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}=-2x+1$$

Add 2x to both sides.

$$8x^{2}+2x=1$$

Subtract 1 from both sides.

$$8x^{2}+2x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 2 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 8\left(-1\right)}}{2\times 8}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 8\left(-1\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-2±\sqrt{4-32\left(-1\right)}}{2\times 8}$$

Multiply -32 times -1.

$$x=\frac{-2±\sqrt{4+32}}{2\times 8}$$

Add 4 to 32.

$$x=\frac{-2±\sqrt{36}}{2\times 8}$$

Take the square root of 36.

$$x=\frac{-2±6}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-2±6}{16}$$

Now solve the equation x=\frac{-2±6}{16} when ± is plus. Add -2 to 6.

$$x=\frac{4}{16}$$

Reduce the fraction \frac{4}{16} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{-2±6}{16} when ± is minus. Subtract 6 from -2.

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

Reduce the fraction \frac{-8}{16} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

$$x=\frac{1}{4} x=-\frac{1}{2}$$

Steps for Completing the Square

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

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

Subtract x^{2} from both sides.

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

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}=-2x+1$$

Add 2x to both sides.

$$8x^{2}+2x=1$$

Divide both sides by 8.

$$\frac{8x^{2}+2x}{8}=\frac{1}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{2}{8}x=\frac{1}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{4}x=\frac{1}{8}$$

Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=\frac{1}{8}+\left(\frac{1}{8}\right)^{2}$$

Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{1}{8}+\frac{1}{64}$$

Add \frac{1}{8} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{9}{64}$$

Factor x^{2}+\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{9}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{9}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{3}{8} x+\frac{1}{8}=-\frac{3}{8}$$

Subtract \frac{1}{8} from both sides of the equation.

$$x=\frac{1}{4} x=-\frac{1}{2}$$