Solve x^2+(7+x)^2=169

$$x ^ { 2 } + ( 7 + x ) ^ { 2 } = 169$$

Solve for x

$$x=-12$$

$$x=5$$

Steps Using Factoring By Grouping

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

$$x^{2}+49+14x+x^{2}=169$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+49+14x=169$$

Subtract 169 from both sides.

$$2x^{2}+49+14x-169=0$$

Subtract 169 from 49 to get -120.

$$2x^{2}-120+14x=0$$

Divide both sides by 2.

$$x^{2}-60+7x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+7x-60=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-60. To find a and b, set up a system to be solved.

$$a+b=7 ab=1\left(-60\right)=-60$$

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 -60.

$$-1,60 -2,30 -3,20 -4,15 -5,12 -6,10$$

Calculate the sum for each pair.

$$-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4$$

The solution is the pair that gives sum 7.

$$a=-5 b=12$$

Rewrite x^{2}+7x-60 as \left(x^{2}-5x\right)+\left(12x-60\right).

$$\left(x^{2}-5x\right)+\left(12x-60\right)$$

Factor out x in the first and 12 in the second group.

$$x\left(x-5\right)+12\left(x-5\right)$$

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

$$\left(x-5\right)\left(x+12\right)$$

To find equation solutions, solve x-5=0 and x+12=0.

$$x=5 x=-12$$

Steps Using the Quadratic Formula

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

$$x^{2}+49+14x+x^{2}=169$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+49+14x=169$$

Subtract 169 from both sides.

$$2x^{2}+49+14x-169=0$$

Subtract 169 from 49 to get -120.

$$2x^{2}-120+14x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}+14x-120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 14 for b, and -120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}-4\times 2\left(-120\right)}}{2\times 2}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 2\left(-120\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-14±\sqrt{196-8\left(-120\right)}}{2\times 2}$$

Multiply -8 times -120.

$$x=\frac{-14±\sqrt{196+960}}{2\times 2}$$

Add 196 to 960.

$$x=\frac{-14±\sqrt{1156}}{2\times 2}$$

Take the square root of 1156.

$$x=\frac{-14±34}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-14±34}{4}$$

Now solve the equation x=\frac{-14±34}{4} when ± is plus. Add -14 to 34.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{-14±34}{4} when ± is minus. Subtract 34 from -14.

$$x=-\frac{48}{4}$$

Divide -48 by 4.

$$x=-12$$

The equation is now solved.

$$x=5 x=-12$$

Steps for Completing the Square

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

$$x^{2}+49+14x+x^{2}=169$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+49+14x=169$$

Subtract 49 from both sides.

$$2x^{2}+14x=169-49$$

Subtract 49 from 169 to get 120.

$$2x^{2}+14x=120$$

Divide both sides by 2.

$$\frac{2x^{2}+14x}{2}=\frac{120}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{14}{2}x=\frac{120}{2}$$

Divide 14 by 2.

$$x^{2}+7x=\frac{120}{2}$$

Divide 120 by 2.

$$x^{2}+7x=60$$

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

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=60+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=60+\frac{49}{4}$$

Add 60 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{289}{4}$$

Factor x^{2}+7x+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{17}{2} x+\frac{7}{2}=-\frac{17}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=5 x=-12$$