SOLUTION: Solve ax^2 + 4a^2 x -12a^3 =0 for x in terms of a by completing the square. Show your steps.

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Question 1001250: Solve ax^2 + 4a^2 x -12a^3 =0 for x in terms of a by completing the square. Show your steps.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
ax%5E2%2B4a%5E2x-12a%5E3=0
a%28x%5E2%2B4ax-12a%5E2%29=0
If a=0 , every real number x is a solution.
If a%3C%3E0 ,
x%5E2%2B4ax-12a%5E2=0
x%5E2%2B4ax=12a%5E2
x%5E2%2B4ax%2B%282a%29%5E2=12a%5E2%2B%282a%29%5E2
%28x%2B2a%29%5E2=12a%5E2%2B4a%5E2
%28x%2B2a%29%5E2=16a%5E2
So,
either x%2B2a=4a--->x=4a-2a--->x=2a ,
or x%2B2a=-4a--->x=-4a-2a--->x=-6a .
You could say highlight%28system%28x=2a%2C%22or%22%2Cx=-6a%29%29 ,
or highlight%28x=-2a+%2B-+4a%29 .