Solve (8x-1)^2=0

$$( 8 x - 1 ) ^ { 2 } = 0$$

Solve for x

$$x=\frac{1}{8}=0.125$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8x-1\right)^{2}.

$$64x^{2}-16x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 64x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-16 ab=64\times 1=64$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 64.

$$-1,-64 -2,-32 -4,-16 -8,-8$$

Calculate the sum for each pair.

$$-1-64=-65 -2-32=-34 -4-16=-20 -8-8=-16$$

The solution is the pair that gives sum -16.

$$a=-8 b=-8$$

Rewrite 64x^{2}-16x+1 as \left(64x^{2}-8x\right)+\left(-8x+1\right).

$$\left(64x^{2}-8x\right)+\left(-8x+1\right)$$

Factor out 8x in the first and -1 in the second group.

$$8x\left(8x-1\right)-\left(8x-1\right)$$

Factor out common term 8x-1 by using distributive property.

$$\left(8x-1\right)\left(8x-1\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve 8x-1=0.

$$x=\frac{1}{8}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8x-1\right)^{2}.

$$64x^{2}-16x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 64 for a, -16 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 64}}{2\times 64}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 64}}{2\times 64}$$

Multiply -4 times 64.

$$x=\frac{-\left(-16\right)±\sqrt{256-256}}{2\times 64}$$

Add 256 to -256.

$$x=\frac{-\left(-16\right)±\sqrt{0}}{2\times 64}$$

Take the square root of 0.

$$x=-\frac{-16}{2\times 64}$$

The opposite of -16 is 16.

$$x=\frac{16}{2\times 64}$$

Multiply 2 times 64.

$$x=\frac{16}{128}$$

Reduce the fraction \frac{16}{128} to lowest terms by extracting and canceling out 16.

$$x=\frac{1}{8}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8x-1\right)^{2}.

$$64x^{2}-16x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$64x^{2}-16x=-1$$

Divide both sides by 64.

$$\frac{64x^{2}-16x}{64}=-\frac{1}{64}$$

Dividing by 64 undoes the multiplication by 64.

$$x^{2}+\left(-\frac{16}{64}\right)x=-\frac{1}{64}$$

Reduce the fraction \frac{-16}{64} to lowest terms by extracting and canceling out 16.

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

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.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=-\frac{1}{64}+\left(-\frac{1}{8}\right)^{2}$$

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

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

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

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=0$$

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}.

$$\left(x-\frac{1}{8}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{1}{8}=0 x-\frac{1}{8}=0$$

Add \frac{1}{8} to both sides of the equation.

$$x=\frac{1}{8} x=\frac{1}{8}$$

The equation is now solved. Solutions are the same.

$$x=\frac{1}{8}$$