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

2 Answers
Jun 20, 2016

Answer:

p = 5 or #-#12

Explanation:

Expand #(p+7)^2#

This expression becomes

#p^2 + (p^2 + 2xxpxx7 + 7^2) = 169#

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

Divide by 2 on both sides

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

#p(p+12) - 5(p+12) = 0#

#(p-5)(p+12) = 0#

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

Hence, p is either 5 or #-#12

Jun 20, 2016

Answer:

#p=5# or p=-12#

Explanation:

Given:

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

#color(white)()#
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 #169# from both sides and transpose to get:

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

Divide both sides by #2# to get:

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

So #p = 5# or #p = -12#

#color(white)()#
Method 2

Note that #169 = 13^2#, so we have:

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