Solve (5x+1)^2=100

$$( 5 x + 1 ) ^ { 2 } = 100$$

Solve for x

$$x = \frac{9}{5} = 1\frac{4}{5} = 1.8$$

$$x = -\frac{11}{5} = -2\frac{1}{5} = -2.2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+1\right)^{2}.

$$25x^{2}+10x+1=100$$

Subtract 100 from both sides.

$$25x^{2}+10x+1-100=0$$

Subtract 100 from 1 to get -99.

$$25x^{2}+10x-99=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25x^{2}+ax+bx-99. To find a and b, set up a system to be solved.

$$a+b=10 ab=25\left(-99\right)=-2475$$

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 -2475.

$$-1,2475 -3,825 -5,495 -9,275 -11,225 -15,165 -25,99 -33,75 -45,55$$

Calculate the sum for each pair.

$$-1+2475=2474 -3+825=822 -5+495=490 -9+275=266 -11+225=214 -15+165=150 -25+99=74 -33+75=42 -45+55=10$$

The solution is the pair that gives sum 10.

$$a=-45 b=55$$

Rewrite 25x^{2}+10x-99 as \left(25x^{2}-45x\right)+\left(55x-99\right).

$$\left(25x^{2}-45x\right)+\left(55x-99\right)$$

Factor out 5x in the first and 11 in the second group.

$$5x\left(5x-9\right)+11\left(5x-9\right)$$

Factor out common term 5x-9 by using distributive property.

$$\left(5x-9\right)\left(5x+11\right)$$

To find equation solutions, solve 5x-9=0 and 5x+11=0.

$$x=\frac{9}{5} x=-\frac{11}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+1\right)^{2}.

$$25x^{2}+10x+1=100$$

Subtract 100 from both sides.

$$25x^{2}+10x+1-100=0$$

Subtract 100 from 1 to get -99.

$$25x^{2}+10x-99=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 10 for b, and -99 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 25\left(-99\right)}}{2\times 25}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 25\left(-99\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-10±\sqrt{100-100\left(-99\right)}}{2\times 25}$$

Multiply -100 times -99.

$$x=\frac{-10±\sqrt{100+9900}}{2\times 25}$$

Add 100 to 9900.

$$x=\frac{-10±\sqrt{10000}}{2\times 25}$$

Take the square root of 10000.

$$x=\frac{-10±100}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-10±100}{50}$$

Now solve the equation x=\frac{-10±100}{50} when ± is plus. Add -10 to 100.

$$x=\frac{90}{50}$$

Reduce the fraction \frac{90}{50} to lowest terms by extracting and canceling out 10.

$$x=\frac{9}{5}$$

Now solve the equation x=\frac{-10±100}{50} when ± is minus. Subtract 100 from -10.

$$x=-\frac{110}{50}$$

Reduce the fraction \frac{-110}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{11}{5}$$

The equation is now solved.

$$x=\frac{9}{5} x=-\frac{11}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+1\right)^{2}.

$$25x^{2}+10x+1=100$$

Subtract 1 from both sides.

$$25x^{2}+10x=100-1$$

Subtract 1 from 100 to get 99.

$$25x^{2}+10x=99$$

Divide both sides by 25.

$$\frac{25x^{2}+10x}{25}=\frac{99}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\frac{10}{25}x=\frac{99}{25}$$

Reduce the fraction \frac{10}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{2}{5}x=\frac{99}{25}$$

Divide \frac{2}{5}, the coefficient of the x term, by 2 to get \frac{1}{5}. Then add the square of \frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{5}x+\left(\frac{1}{5}\right)^{2}=\frac{99}{25}+\left(\frac{1}{5}\right)^{2}$$

Square \frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{99+1}{25}$$

Add \frac{99}{25} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=4$$

Factor x^{2}+\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{5}\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{5}\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+\frac{1}{5}=2 x+\frac{1}{5}=-2$$

Subtract \frac{1}{5} from both sides of the equation.

$$x=\frac{9}{5} x=-\frac{11}{5}$$