Solve (16m-1)^2=0

$$( 16 m - 1 ) ^ { 2 } = 0$$

Solve for m

$$m=\frac{1}{16}=0.0625$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(16m-1\right)^{2}.

$$256m^{2}-32m+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 256m^{2}+am+bm+1. To find a and b, set up a system to be solved.

$$a+b=-32 ab=256\times 1=256$$

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 256.

$$-1,-256 -2,-128 -4,-64 -8,-32 -16,-16$$

Calculate the sum for each pair.

$$-1-256=-257 -2-128=-130 -4-64=-68 -8-32=-40 -16-16=-32$$

The solution is the pair that gives sum -32.

$$a=-16 b=-16$$

Rewrite 256m^{2}-32m+1 as \left(256m^{2}-16m\right)+\left(-16m+1\right).

$$\left(256m^{2}-16m\right)+\left(-16m+1\right)$$

Factor out 16m in the first and -1 in the second group.

$$16m\left(16m-1\right)-\left(16m-1\right)$$

Factor out common term 16m-1 by using distributive property.

$$\left(16m-1\right)\left(16m-1\right)$$

Rewrite as a binomial square.

$$\left(16m-1\right)^{2}$$

To find equation solution, solve 16m-1=0.

$$m=\frac{1}{16}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(16m-1\right)^{2}.

$$256m^{2}-32m+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 256 for a, -32 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 256}}{2\times 256}$$

Square -32.

$$m=\frac{-\left(-32\right)±\sqrt{1024-4\times 256}}{2\times 256}$$

Multiply -4 times 256.

$$m=\frac{-\left(-32\right)±\sqrt{1024-1024}}{2\times 256}$$

Add 1024 to -1024.

$$m=\frac{-\left(-32\right)±\sqrt{0}}{2\times 256}$$

Take the square root of 0.

$$m=-\frac{-32}{2\times 256}$$

The opposite of -32 is 32.

$$m=\frac{32}{2\times 256}$$

Multiply 2 times 256.

$$m=\frac{32}{512}$$

Reduce the fraction \frac{32}{512} to lowest terms by extracting and canceling out 32.

$$m=\frac{1}{16}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(16m-1\right)^{2}.

$$256m^{2}-32m+1=0$$

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

$$256m^{2}-32m=-1$$

Divide both sides by 256.

$$\frac{256m^{2}-32m}{256}=-\frac{1}{256}$$

Dividing by 256 undoes the multiplication by 256.

$$m^{2}+\left(-\frac{32}{256}\right)m=-\frac{1}{256}$$

Reduce the fraction \frac{-32}{256} to lowest terms by extracting and canceling out 32.

$$m^{2}-\frac{1}{8}m=-\frac{1}{256}$$

Divide -\frac{1}{8}, the coefficient of the x term, by 2 to get -\frac{1}{16}. Then add the square of -\frac{1}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}-\frac{1}{8}m+\left(-\frac{1}{16}\right)^{2}=-\frac{1}{256}+\left(-\frac{1}{16}\right)^{2}$$

Square -\frac{1}{16} by squaring both the numerator and the denominator of the fraction.

$$m^{2}-\frac{1}{8}m+\frac{1}{256}=\frac{-1+1}{256}$$

Add -\frac{1}{256} to \frac{1}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$m^{2}-\frac{1}{8}m+\frac{1}{256}=0$$

Factor m^{2}-\frac{1}{8}m+\frac{1}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(m-\frac{1}{16}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-\frac{1}{16}\right)^{2}}=\sqrt{0}$$

Simplify.

$$m-\frac{1}{16}=0 m-\frac{1}{16}=0$$

Add \frac{1}{16} to both sides of the equation.

$$m=\frac{1}{16} m=\frac{1}{16}$$

The equation is now solved. Solutions are the same.

$$m=\frac{1}{16}$$