Solve (7m-1)^2-100=0

$$( 7 m - 1 ) ^ { 2 } - 100 = 0$$

Solve for m

$$m = \frac{11}{7} = 1\frac{4}{7} \approx 1.571428571$$

$$m = -\frac{9}{7} = -1\frac{2}{7} \approx -1.285714286$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7m-1\right)^{2}.

$$49m^{2}-14m+1-100=0$$

Subtract 100 from 1 to get -99.

$$49m^{2}-14m-99=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 49m^{2}+am+bm-99. To find a and b, set up a system to be solved.

$$a+b=-14 ab=49\left(-99\right)=-4851$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -4851.

$$1,-4851 3,-1617 7,-693 9,-539 11,-441 21,-231 33,-147 49,-99 63,-77$$

Calculate the sum for each pair.

$$1-4851=-4850 3-1617=-1614 7-693=-686 9-539=-530 11-441=-430 21-231=-210 33-147=-114 49-99=-50 63-77=-14$$

The solution is the pair that gives sum -14.

$$a=-77 b=63$$

Rewrite 49m^{2}-14m-99 as \left(49m^{2}-77m\right)+\left(63m-99\right).

$$\left(49m^{2}-77m\right)+\left(63m-99\right)$$

Factor out 7m in the first and 9 in the second group.

$$7m\left(7m-11\right)+9\left(7m-11\right)$$

Factor out common term 7m-11 by using distributive property.

$$\left(7m-11\right)\left(7m+9\right)$$

To find equation solutions, solve 7m-11=0 and 7m+9=0.

$$m=\frac{11}{7} m=-\frac{9}{7}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7m-1\right)^{2}.

$$49m^{2}-14m+1-100=0$$

Subtract 100 from 1 to get -99.

$$49m^{2}-14m-99=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 49 for a, -14 for b, and -99 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 49\left(-99\right)}}{2\times 49}$$

Square -14.

$$m=\frac{-\left(-14\right)±\sqrt{196-4\times 49\left(-99\right)}}{2\times 49}$$

Multiply -4 times 49.

$$m=\frac{-\left(-14\right)±\sqrt{196-196\left(-99\right)}}{2\times 49}$$

Multiply -196 times -99.

$$m=\frac{-\left(-14\right)±\sqrt{196+19404}}{2\times 49}$$

Add 196 to 19404.

$$m=\frac{-\left(-14\right)±\sqrt{19600}}{2\times 49}$$

Take the square root of 19600.

$$m=\frac{-\left(-14\right)±140}{2\times 49}$$

The opposite of -14 is 14.

$$m=\frac{14±140}{2\times 49}$$

Multiply 2 times 49.

$$m=\frac{14±140}{98}$$

Now solve the equation m=\frac{14±140}{98} when ± is plus. Add 14 to 140.

$$m=\frac{154}{98}$$

Reduce the fraction \frac{154}{98} to lowest terms by extracting and canceling out 14.

$$m=\frac{11}{7}$$

Now solve the equation m=\frac{14±140}{98} when ± is minus. Subtract 140 from 14.

$$m=-\frac{126}{98}$$

Reduce the fraction \frac{-126}{98} to lowest terms by extracting and canceling out 14.

$$m=-\frac{9}{7}$$

The equation is now solved.

$$m=\frac{11}{7} m=-\frac{9}{7}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7m-1\right)^{2}.

$$49m^{2}-14m+1-100=0$$

Subtract 100 from 1 to get -99.

$$49m^{2}-14m-99=0$$

Add 99 to both sides. Anything plus zero gives itself.

$$49m^{2}-14m=99$$

Divide both sides by 49.

$$\frac{49m^{2}-14m}{49}=\frac{99}{49}$$

Dividing by 49 undoes the multiplication by 49.

$$m^{2}+\left(-\frac{14}{49}\right)m=\frac{99}{49}$$

Reduce the fraction \frac{-14}{49} to lowest terms by extracting and canceling out 7.

$$m^{2}-\frac{2}{7}m=\frac{99}{49}$$

Divide -\frac{2}{7}, the coefficient of the x term, by 2 to get -\frac{1}{7}. Then add the square of -\frac{1}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}-\frac{2}{7}m+\left(-\frac{1}{7}\right)^{2}=\frac{99}{49}+\left(-\frac{1}{7}\right)^{2}$$

Square -\frac{1}{7} by squaring both the numerator and the denominator of the fraction.

$$m^{2}-\frac{2}{7}m+\frac{1}{49}=\frac{99+1}{49}$$

Add \frac{99}{49} to \frac{1}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$m^{2}-\frac{2}{7}m+\frac{1}{49}=\frac{100}{49}$$

Factor m^{2}-\frac{2}{7}m+\frac{1}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(m-\frac{1}{7}\right)^{2}=\frac{100}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-\frac{1}{7}\right)^{2}}=\sqrt{\frac{100}{49}}$$

Simplify.

$$m-\frac{1}{7}=\frac{10}{7} m-\frac{1}{7}=-\frac{10}{7}$$

Add \frac{1}{7} to both sides of the equation.

$$m=\frac{11}{7} m=-\frac{9}{7}$$