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

##### 2 Answers

p = 5 or

#### Explanation:

Expand

This expression becomes

Divide by 2 on both sides

Now choose two numbers such that their sum is coefficient of p i.e.

Such numbers are

So,

So either

Hence, p is either 5 or

#### Explanation:

Given:

#p^2+(p+7)^2 = 169#

**Method 1**

Rearrange into standard polynomial form as follows:

#169 = p^2+(p+7)^2#

#=p^2+p^2+14p+49#

#=2p^2+14p+49#

Subtract

#2p^2+14p-120 = 0#

Divide both sides by

#p^2+7p-60 = 0#

To solve this, we can look for a pair of factors of

#0 = p^2+7p-60 = (p+12)(p-5)#

So

**Method 2**

Note that

#p^2+(p+7)^2 = 13^2#

This is in the form

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

Note that we have a quadratic equation, so it will have a second root. What could that be? Putting

#-12, -5, 13#

So the other root is