Question

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

Answer 1

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

Answer 2

`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`