Solve (13x+1)^2=64

$$( 13 x + 1 ) ^ { 2 } = 64$$

Solve for x

$$x=\frac{7}{13}\approx 0.538461538$$

$$x=-\frac{9}{13}\approx -0.692307692$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(13x+1\right)^{2}.

$$169x^{2}+26x+1=64$$

Subtract 64 from both sides.

$$169x^{2}+26x+1-64=0$$

Subtract 64 from 1 to get -63.

$$169x^{2}+26x-63=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 169x^{2}+ax+bx-63. To find a and b, set up a system to be solved.

$$a+b=26 ab=169\left(-63\right)=-10647$$

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 -10647.

$$-1,10647 -3,3549 -7,1521 -9,1183 -13,819 -21,507 -39,273 -63,169 -91,117$$

Calculate the sum for each pair.

$$-1+10647=10646 -3+3549=3546 -7+1521=1514 -9+1183=1174 -13+819=806 -21+507=486 -39+273=234 -63+169=106 -91+117=26$$

The solution is the pair that gives sum 26.

$$a=-91 b=117$$

Rewrite 169x^{2}+26x-63 as \left(169x^{2}-91x\right)+\left(117x-63\right).

$$\left(169x^{2}-91x\right)+\left(117x-63\right)$$

Factor out 13x in the first and 9 in the second group.

$$13x\left(13x-7\right)+9\left(13x-7\right)$$

Factor out common term 13x-7 by using distributive property.

$$\left(13x-7\right)\left(13x+9\right)$$

To find equation solutions, solve 13x-7=0 and 13x+9=0.

$$x=\frac{7}{13} x=-\frac{9}{13}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(13x+1\right)^{2}.

$$169x^{2}+26x+1=64$$

Subtract 64 from both sides.

$$169x^{2}+26x+1-64=0$$

Subtract 64 from 1 to get -63.

$$169x^{2}+26x-63=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 169 for a, 26 for b, and -63 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-26±\sqrt{26^{2}-4\times 169\left(-63\right)}}{2\times 169}$$

Square 26.

$$x=\frac{-26±\sqrt{676-4\times 169\left(-63\right)}}{2\times 169}$$

Multiply -4 times 169.

$$x=\frac{-26±\sqrt{676-676\left(-63\right)}}{2\times 169}$$

Multiply -676 times -63.

$$x=\frac{-26±\sqrt{676+42588}}{2\times 169}$$

Add 676 to 42588.

$$x=\frac{-26±\sqrt{43264}}{2\times 169}$$

Take the square root of 43264.

$$x=\frac{-26±208}{2\times 169}$$

Multiply 2 times 169.

$$x=\frac{-26±208}{338}$$

Now solve the equation x=\frac{-26±208}{338} when ± is plus. Add -26 to 208.

$$x=\frac{182}{338}$$

Reduce the fraction \frac{182}{338} to lowest terms by extracting and canceling out 26.

$$x=\frac{7}{13}$$

Now solve the equation x=\frac{-26±208}{338} when ± is minus. Subtract 208 from -26.

$$x=-\frac{234}{338}$$

Reduce the fraction \frac{-234}{338} to lowest terms by extracting and canceling out 26.

$$x=-\frac{9}{13}$$

The equation is now solved.

$$x=\frac{7}{13} x=-\frac{9}{13}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(13x+1\right)^{2}.

$$169x^{2}+26x+1=64$$

Subtract 1 from both sides.

$$169x^{2}+26x=64-1$$

Subtract 1 from 64 to get 63.

$$169x^{2}+26x=63$$

Divide both sides by 169.

$$\frac{169x^{2}+26x}{169}=\frac{63}{169}$$

Dividing by 169 undoes the multiplication by 169.

$$x^{2}+\frac{26}{169}x=\frac{63}{169}$$

Reduce the fraction \frac{26}{169} to lowest terms by extracting and canceling out 13.

$$x^{2}+\frac{2}{13}x=\frac{63}{169}$$

Divide \frac{2}{13}, the coefficient of the x term, by 2 to get \frac{1}{13}. Then add the square of \frac{1}{13} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{13}x+\left(\frac{1}{13}\right)^{2}=\frac{63}{169}+\left(\frac{1}{13}\right)^{2}$$

Square \frac{1}{13} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{13}x+\frac{1}{169}=\frac{63+1}{169}$$

Add \frac{63}{169} to \frac{1}{169} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{13}x+\frac{1}{169}=\frac{64}{169}$$

Factor x^{2}+\frac{2}{13}x+\frac{1}{169}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{13}\right)^{2}=\frac{64}{169}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{13}\right)^{2}}=\sqrt{\frac{64}{169}}$$

Simplify.

$$x+\frac{1}{13}=\frac{8}{13} x+\frac{1}{13}=-\frac{8}{13}$$

Subtract \frac{1}{13} from both sides of the equation.

$$x=\frac{7}{13} x=-\frac{9}{13}$$