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

2 Answers
Jun 20, 2016

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

p=5 or p=-12#

Explanation:

Given:

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

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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

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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