SOLUTION: p^2+q^2=85, p-q=1,then p^2-q^2=?

Algebra ->  Test -> SOLUTION: p^2+q^2=85, p-q=1,then p^2-q^2=?      Log On


   

Question 759039: p^2+q^2=85, p-q=1,then p^2-q^2=?
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
p^2+q^2=85, p-q=1,then p^2-q^2=?
-------
p = 1+q
-----
Equation:
(1+q)^2 + q^2 = 85
1 + 2q + 2q^2 = 85
----------------------------------
2q^2 + 2q - 84 = 0
------
q^2 + q - 42 = 0
----
Factor:
(q+7)(q-6) = 0
---------------------
q = 6 or q = -7
----
Solve for "p"
p - q = 1
If q = 6, p = 7
If q = -7, p = -6
-----
p^2-q^2=?

P^2 - q^2 = 6^2-7^2 = -13
OR
p^2 - q^2 = 7^2 - 6^2 = 13
========================
Cheers,
Stan H.
====================

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

p%5E2%2Bq%5E2=85,
if p-q=1,then p=q%2B1........plug it in p%5E2%2Bq%5E2=85
%28q%2B1%29%5E2%2Bq%5E2=85
q%5E2%2B2q%2B1%2Bq%5E2=85
2q%5E2%2B2q%2B1=85
2q%5E2%2B2q=84...divide all by 2
q%5E2%2Bq=42
q%5E2%2Bq-42=0......use quadratic formula

q+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
q+=+%28-1+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-42%29+%29%29%2F%282%2A1%29+
q+=+%28-1+%2B-+sqrt%28+1%2B168+%29%29%2F2+
q+=+%28-1+%2B-+sqrt%28169+%29%29%2F2+
q+=+%28-1+%2B-+13%29%2F2+
solutions:
q+=+%28-1+%2B+13%29%2F2+
q+=+12%2F2+
highlight%28q=6%29
or
q+=+%28-1+-+13%29%2F2+
q+=+-14%2F2+
highlight%28q=-7%29


now find p
p-q=1
p-6=1
p=1%2B6
highlight%28p=7%29
or
p-%28-7%29=1
p%2B7=1
p=1-7
highlight%28p=-6%29
so, there are two solutions:
1. highlight%28p=7%29 and highlight%28q=6%29
2. highlight%28p=-7%29 and highlight%28q=-6%29

then
p%5E2-q%5E2=7%5E2-6%5E2=49-36=13
or
p%5E2-q%5E2=%28-7%29%5E2-%28-6%29%5E2=49-36=13