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

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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) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
Let the numbers be:
+x+-+5+
+x+
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+x%5E2+%2B+%28+x-5+%29%5E2+=+97+
+x%5E2+%2B+x%5E2+-+10x+%2B+25+=+97+
+2x%5E2+-+10x+-+72+=+0+
+x%5E2+-+5x+-+36+=+0+
+%28+x+-+9+%29%2A%28+x+%2B+4+%29+=+0+ ( by looking at it )
+x+=+9+
+x+-+5+=+4+
( I will use the positive result )
The numbers are 9 and 4
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check:
+x%5E2+%2B+%28+x-5+%29%5E2+=+97+
+9%5E2+%2B+%28+9-5+%29%5E2+=+97+
+81+%2B+16+=+97+
+97+=+97+
OK
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The negative answers also work:
+x+%2B+4+=+0+
+x+=+-4+
and
+x+-+5+=+-9+
The numbers are -4 and -9
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