SOLUTION: I just learned how to solve quadratic equations by completing the square and my book throws this doozy in a section called "Challenge", can this be solved by completing the square?

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Question 1083261: I just learned how to solve quadratic equations by completing the square and my book throws this doozy in a section called "Challenge", can this be solved by completing the square? Are there more than one way to solve it?
Solve for x in terms of a.
3x^2+ax^2=9x+9a
Found 2 solutions by Alan3354, KMST:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
3x^2+ax^2=9x+9a
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3x^2 + ax^2 - 9x = 9a
(a+3)*x^2 - 9x = 9a
x%5E2+-+x%2A%289%2F%28a%2B3%29%29+=+9a%2F%28a%2B3%29
b = 9/(a+3)
Add (9/2(a+3))^2 = (81/4)/(a+3)^2
----

%28x+-+9%2F%282%28a%2B3%29%29%29%5E2+=+9a%2F%28a%2B3%29+%2B+81%2F%284%28a%2B3%29%5E2%29

%28x+-+9%2F%282%28a%2B3%29%29%29%5E2+=+%2836a%5E2+%2B108a+%2B+81%29%2F%284%28a%2B3%29%5E2%29

%28x+-+9%2F%282%28a%2B3%29%29%29%5E2+=+9%2A%282a+%2B+3%29%5E2%2F%284%28a%2B3%29%5E2%29
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x+-+18%2F%284%28a%2B3%29%29+=+%2B%286a+%2B+9%29%2F%284%28a%2B3%29%29
x+=+%286a+%2B+27%29%2F%284%28a%2B3%29%29
=====================================
x+-+18%2F%284%28a%2B3%29%29+=+-%286a+%2B+9%29%2F%284%28a%2B3%29%29
x+=+%28-6a+%2B+9%29%2F%284%28a%2B3%29%29
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Very tedious. Mistakes are possible.
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Any quadratic with real number coeffici3ents can be solved by completing the square.
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Using the quadratic equation will work, too.
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I like the other tutor's approach and solution better.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Some (not all) quadratic equation can be solved by factoring.
Every quadratic equation can be solved by completing the square,
and every quadratic equation can be solved by using the quadratic formula.
People usually resort to the quadratic formula,
but professor Peter Alfeld of the university of Utah
believes we should complete squares.
I use what comes easier. When factoring works, it may be easier.

3x%5E2%2Bax%5E2=9x%2B9a
%283%2Ba%29x%5E2-9x=9a
Maybe dividing everything by %283%2Ba%29 at this point would be easier.
Of course, we will have to add a%3C%3E-3 as a requirement to any solution we find as a function of a.
x%5E2-%289%2F%283%2Ba%29%29x=9a%2F%283%2Ba%29

%28x-9%2F%282a%2B6%29%29%5E2=9a%2F%283%2Ba%29%2B%289%2F%282%283%2Ba%29%29%29%5E2
%28x-9%2F%282a%2B6%29%29%5E2%22=%229a%284%28a%2B3%29%29%2F%282a%2B6%29%5E2%22%2B%229%5E2%2F%282a%2B6%29%5E2
%28x-9%2F%282a%2B6%29%29%5E2%22=%22%289%284a%5E2%2B12%29%2B81%29%2F%282a%2B6%29%5E2
%28x-9%2F%282a%2B6%29%29%5E2%22=%229%284a%5E2%2B12a%2B9%29%2F%282a%2B6%29%5E2
%28x-9%2F%282a%2B6%29%29%5E2%22=%229%282a%2B3%29%5E2%2F%282a%2B6%29%5E2
x=9%2F%282a%2B6%29+%2B-+3%282a%2B3%29%2F%282a%2B6%29
x=%289+%2B-+%286a%2B9%29%29%2F%282a%2B6%29
So, the two solutions are
x%5B1%5D=%286a%2B18%29%2F%282a%2B6%29=3 and x%5B2%5D=6a%2F%282a%2B6%29=3a%2F%28a%2B3%29 if a%3C%3E-3

WAIT!
Why did we not realize that we could solve by factoring?
3x%5E2%2Bax%5E2=9x%2B9a
ax%5E2-9a%2B3x%5E2-9x=0
3x%28x-3%29=a%289-x%5E2%29
a%28x%5E2-9%29%2B3x%28x-3%29=0
a%28x%2B3%29%28x-3%29%2B3x%28x-3%29=0
%28a%28x%2B3%29%2B3x%29%28x-3%29=0
%28ax%2B3a%2B3x%29%28x-3%29=0
%28%28a%2B3%29x-3a%29%28x-3%29=0
At this point, either x-3=0 ---> x=3 , or
%28a%2B3%29x-3a=0 ---> x=3a%2F%28a%2B3%29 as long as a%3C%3E-3 .