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Math Problem

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Re: Math Problem

by billzhao » Mon Feb 16, 2009 5:30 pm
trojansc82 wrote:How do I solve for a in terms of b?

((a^2)/(b^2)) + (2a/b) = -1
Multiply the both sides with b^2 =>

a^2+2*a*b=-b^2 =>

a^2+2*a*b+b^2=0 =>

(a+b)^2=0 =>

a+b=0 =>

a=-b
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by AJ2009 » Tue Feb 17, 2009 9:33 am
if you multiply both sides with b^2 you will get a^2+2ab^3=-b^2 and not a^2+2ab=-b^2.

this solution will only work if the expression is (a^2+2ab)/b^2 = -1.
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by sureshbala » Tue Feb 17, 2009 11:02 am
Let a/b = x.

Given x^2 + 2x = -1

i.e x^2 + 2x + 1 =0

i.e (x+1)^2 = 0

i.e x = -1

So a/b = -1 i.e a = -b
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by marcusking » Tue Feb 17, 2009 11:09 am
AJ2009 wrote:if you multiply both sides with b^2 you will get a^2+2ab^3=-b^2 and not a^2+2ab=-b^2.

this solution will only work if the expression is (a^2+2ab)/b^2 = -1.
Check your work again. It may help to get everything except a^2/b^2 on one side.

a^2 = (-b^2) - (2a /b) b^2
a^2 = (-b^2) - (2a*b^2)/ b
a^2 = (-b^2) - (2ab)
a^2 + b^2 + 2ab = 0;
(a+b)^2 = 0
a = -b
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by ven4gmat » Tue Feb 17, 2009 12:40 pm
sureshbala wrote:Let a/b = x.

Given x^2 + 2x = -1

i.e x^2 + 2x + 1 =0

i.e (x+1)^2 = 0

i.e x = -1

So a/b = -1 i.e a = -b
This will be my way of doing in the exam....simple and quick
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