Solve a^2+(7-a)^2=169

$$a ^ { 2 } + ( 7 - a ) ^ { 2 } = 169$$

Solve for a

$$a=-5$$

$$a=12$$

Steps Using Factoring By Grouping

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

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

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

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

Subtract 169 from both sides.

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

Subtract 169 from 49 to get -120.

$$2a^{2}-120-14a=0$$

Divide both sides by 2.

$$a^{2}-60-7a=0$$

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

$$a^{2}-7a-60=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba-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 negative, the negative number has greater absolute value than the positive. 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.

$$a=-12 b=5$$

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

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

Factor out a in the first and 5 in the second group.

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

Factor out common term a-12 by using distributive property.

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

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

$$a=12 a=-5$$

Steps Using the Quadratic Formula

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

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

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

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

Subtract 169 from both sides.

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

Subtract 169 from 49 to get -120.

$$2a^{2}-120-14a=0$$

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.

$$2a^{2}-14a-120=0$$

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}.

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

Square -14.

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

Multiply -4 times 2.

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

Multiply -8 times -120.

$$a=\frac{-\left(-14\right)±\sqrt{196+960}}{2\times 2}$$

Add 196 to 960.

$$a=\frac{-\left(-14\right)±\sqrt{1156}}{2\times 2}$$

Take the square root of 1156.

$$a=\frac{-\left(-14\right)±34}{2\times 2}$$

The opposite of -14 is 14.

$$a=\frac{14±34}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{14±34}{4}$$

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

$$a=\frac{48}{4}$$

Divide 48 by 4.

$$a=12$$

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

$$a=-\frac{20}{4}$$

Divide -20 by 4.

$$a=-5$$

The equation is now solved.

$$a=12 a=-5$$

Steps for Completing the Square

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

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

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

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

Subtract 49 from both sides.

$$2a^{2}-14a=169-49$$

Subtract 49 from 169 to get 120.

$$2a^{2}-14a=120$$

Divide both sides by 2.

$$\frac{2a^{2}-14a}{2}=\frac{120}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$a^{2}+\left(-\frac{14}{2}\right)a=\frac{120}{2}$$

Divide -14 by 2.

$$a^{2}-7a=\frac{120}{2}$$

Divide 120 by 2.

$$a^{2}-7a=60$$

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.

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

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

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

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

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

Factor a^{2}-7a+\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}.

$$\left(a-\frac{7}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

$$a-\frac{7}{2}=\frac{17}{2} a-\frac{7}{2}=-\frac{17}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$a=12 a=-5$$