Thread: [SOLVED] Quadratic Equation
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Old May 2nd, 2008, 06:42 AM
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Quote:
Originally Posted by looi76 View Post
Question:
Find the value of x when y = 0
y = \frac{2x - 1}{x + 3}

Attempt:

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

(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?
You've made very heavy weather of this. Simply solve 2x - 1 - 0 (why?).

And it's good form to check that the solution is not a solution to x + 3 = 0 (why?)
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