SOLUTION: A complicated question: "Find both roots of the following equation to the nearest tenth. (a+3)(squared)+(a-3)(squared)=256."

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Question 250852: A complicated question: "Find both roots of the following equation to the nearest tenth. (a+3)(squared)+(a-3)(squared)=256."
Found 2 solutions by College Student, ankor@dixie-net.com:
Answer by College Student(505) About Me  (Show Source):
You can put this solution on YOUR website!
Problem: %28%28a%2B3%29%5E2%29%2B%28%28a-3%29%5E2%29=256

If we take the square root of the entire equation, we get:

%28a%2B3%29%2B%28a-3%29=16

Simplify:
2a=16

a=16%2F2

a=8 Done!

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
"Find both roots of the following equation to the nearest tenth.
(a+3)(squared)+(a-3)(squared)=256."
:
%28a%2B3%29%5E2+%2B+%28a-3%29%5E2 = 256
FOIL
a%5E2+%2B+6a+%2B+9+%2B+a%5E2+-+6a+%2B+9+=+256
Combine like terms
a%5E2+%2B+a%5E2+%2B+6a+-+6a+%2B+9+%2B+9+=+256
2a%5E2+%2B+18+=+256
2a%5E2+=+256+-+18
2a%5E2+=+238
a%5E2+=+238%2F2
a%5E2+=+119
a = +/-sqrt%28119%29
a = +10.9
a = -10.9
;
:
Check solution in original equation
13.9^2 + 7.9^2 = 255.6 ~ 256