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