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Moo · #2 May 2nd, 2008, 06:42 AM

Hello,

Quote:

Originally Posted by looi76

Question:
Find the value of $x$ when $y = 0$
$y = \frac{2x - 1}{x + 3}$

Attempt:

$(2x - 1)(x + 3)^{-1} = 0$

${\color{red}(2x - 1)(x^{-1} + \frac{1}{3}) = 0}$

$2 + \frac{2}{3}x - x^{-1} - \frac{1}{3} = 0$

$\frac{2}{3}x - x^{-1} + \frac{5}{3} = 0$

$\frac{4}{9}x^2 - 1 + \frac{5}{3} = 0$

$\frac{4}{9}x^2 + \frac{5}{3} - 1 = 0$

I think the quadratic equation that i got is wrong

... Where did I go wrong?

It's the step in red which is incorrect...

$(a+b)^n \not = a^n+b^n$

What you have to do is to say that x+3 has to be different of 0.
Then solve :

$0=\frac{2x-1}{x+3}$

Multiplying both sides by x+3 (which is possible because x+3 is not = 0) :

$\boxed{0=2x-1}$

But I don't see why there is a quadratic equation here... ^^

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looi76
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