Question 1083261
3x^2+ax^2=9x+9a
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3x^2 + ax^2 - 9x = 9a
(a+3)*x^2 - 9x = 9a
{{{x^2 - x*(9/(a+3)) = 9a/(a+3)}}}
b = 9/(a+3)
Add (9/2(a+3))^2 = (81/4)/(a+3)^2
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{{{x^2 - x*(9/(a+3)) + (81/4)/(a+3)^2 = 9a/(a+3) + (81/4)/(a+3)^2}}}
{{{(x - 9/(2(a+3)))^2 = 9a/(a+3) + 81/(4(a+3)^2)}}}
{{{(x - 9/(2(a+3)))^2 = 9a*4(a+3)/(4(a+3)^2) + 81/(4(a+3)^2)}}}
{{{(x - 9/(2(a+3)))^2 = (36a^2 +108a + 81)/(4(a+3)^2)}}}
{{{(x - 9/(2(a+3)))^2 = 9*(4a^2 + 12a + 9)/(4(a+3)^2)}}}
{{{(x - 9/(2(a+3)))^2 = 9*(2a + 3)^2/(4(a+3)^2)}}}
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{{{x - 18/(4(a+3)) = +(6a + 9)/(4(a+3))}}}
{{{x = (6a + 27)/(4(a+3))}}}
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{{{x - 18/(4(a+3)) = -(6a + 9)/(4(a+3))}}}
{{{x = (-6a + 9)/(4(a+3))}}}
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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.