SOLUTION: One number is five less than another. The sum of their squares is 97. Find the numbers.

Question 1144836: One number is five less than another. The sum of their squares is 97. Find the numbers.
Found 2 solutions by ankor@dixie-net.com, josmiceli:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
One number is five less than another.
a = b - 5
The sum of their squares is 97.
a^2 + b^2 = 97
Find the numbers.
replace a with (b-5)
(b-5)^2 + b^2 = 97
FOIL (b-5)(b-5)
b^2 - 5b - 5b + 25 + b^2 = 97
Combine like terms
2b^2 - 10b + 25 - 97 = 0
A quadratic equation
2b^2 - 10b - 72 = =
simplify, divide by 2
b^2 - 5b - 36 = 0
Factors to
(b - 9)(b + 4) = 0
the positive solution
b = 9
and
a = 9 - 5
a = = 4
:
The numbers are 9 & 4
:
However, b = -4, is another solution, therefore -9 and -4 can also be the numbers
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let the numbers be:

---------------

( by looking at it )

( I will use the positive result )
The numbers are 9 and 4
-----------------------------------
check:

OK
---------------------------
The negative answers also work:

and

The numbers are -4 and -9
--------------------------------

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