# How do you solve p^2+(p+7)^2=169?

Jun 20, 2016

p = 5 or $-$12

#### Explanation:

Expand ${\left(p + 7\right)}^{2}$

This expression becomes

${p}^{2} + \left({p}^{2} + 2 \times p \times 7 + {7}^{2}\right) = 169$

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

Divide by 2 on both sides

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

Now choose two numbers such that their sum is coefficient of p i.e. $- 7$ and product is constant term i.e. $- 60$.

Such numbers are $12$ and $- 5$

So,

${p}^{2} + 12 p - 5 p - 60 = 0$

$p \left(p + 12\right) - 5 \left(p + 12\right) = 0$

$\left(p - 5\right) \left(p + 12\right) = 0$

So either $p - 5 = 0$ or $p + 12 = 0$

Hence, p is either 5 or $-$12

Jun 20, 2016

$p = 5$ or p=-12#

#### Explanation:

Given:

${p}^{2} + {\left(p + 7\right)}^{2} = 169$

$\textcolor{w h i t e}{}$
Method 1

Rearrange into standard polynomial form as follows:

$169 = {p}^{2} + {\left(p + 7\right)}^{2}$

$= {p}^{2} + {p}^{2} + 14 p + 49$

$= 2 {p}^{2} + 14 p + 49$

Subtract $169$ from both sides and transpose to get:

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

Divide both sides by $2$ to get:

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

To solve this, we can look for a pair of factors of $60$ with difference $7$. The pair $12 , 5$ works, so we have:

$0 = {p}^{2} + 7 p - 60 = \left(p + 12\right) \left(p - 5\right)$

So $p = 5$ or $p = - 12$

$\textcolor{w h i t e}{}$
Method 2

Note that $169 = {13}^{2}$, so we have:

${p}^{2} + {\left(p + 7\right)}^{2} = {13}^{2}$

This is in the form ${a}^{2} + {b}^{2} = {c}^{2}$

So we are looking for a Pythagorean triple:

$p , p + 7 , 13$

The first couple of positive Pythagorean triples that are not scalar multiples of smaller ones are:

$3 , 4 , 5$

$5 , 12 , 13$

The second one matches, so we find a solution $p = 5$.

Note that we have a quadratic equation, so it will have a second root. What could that be? Putting $- 12$ into our equation, we find that gives us the Pythagorean triple:

$- 12 , - 5 , 13$

So the other root is $p = - 12$