Question 940261: One number is six more than another number, the sum of their squares is 90?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! let the two numbers be a & b
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Write an equation for each statement
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One number is six more than another number,
a = b + 6
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the sum of their squares is 90?
a^2 + b^2 = 90
Replace a with (b+6)
(b+6)^2 + b^2 = 90
FOIL (b+6)(b+6)
b^2 + 6b + 6b + 36 + b^2 = 90
Combine like terms, write as a quadratic equation
b^2 + b^2 + 12b + 36 - 90 = 0
2b^2 + 12b - 54 = 0
simplify, divide by 2
b^2 + 6b - 27 = 0
Factors to
(b+9)(b-3) = 0
two solutions
b = -9
b = +3
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find a when b = -9
a - (-9) = 6
a + 9 = 6
a = 6 - 9
a = -3
find a when b = 3
a - 3 = 6
a = 6 + 3
a = 9
:
Summarize: a=-3; b=-9. Or a=9; b=3
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