# Thread: [SOLVED] solve the equation

1. ## [SOLVED] solve the equation

Can someone show me how to solve this equation please? This is part of something bigger, but what I'm stuck on is just algebra
$\displaystyle 2x^2+x-1=0$

I can distribute it to x(2x+1) = 1, but am not really sure what to do from there, or if that's even the proper way to go. I would normally try to factor, but I don't think this is factorable. Any assistance is appreciated.

2. I think it does factorize to give $\displaystyle (2x-1)(x+1)=0$

3. you're right, it does factor. I was totally unable to figure that out. Thank you for the assistance.

4. There is a test you can use to see if something can be factorized:

For any quadratic, $\displaystyle ax^2+bx+c=0$,
Take the discriminant $\displaystyle b^2-4(a)(c)$
If the result is a perfect square (i.e. it can be square-rooted and give an integer result), then the sum in question can be factorized.

$\displaystyle b^2-4(a)(c)$
$\displaystyle 1^2-4(2)(-1)$
$\displaystyle =1+8$
$\displaystyle =9$
This is a perfect square so the quadratic can be factorized.

5. Are You sure, You want solve : $\displaystyle 2x^2+x-1=0$ ??

So, $\displaystyle 2x^2+x-1=0$ is a equation the second degré

$\displaystyle \Delta = b^2-4ac = 1^2 - 4*2*(-1) = 9 \geq 0$

The Solutions : $\displaystyle x_1= \frac{-1+ \sqrt{9}}{4} = \frac{1}{2}$ , $\displaystyle x_2 = \frac{-1-\sqrt{9}}{4} = -1$

6. Thanks Quacky, that's very instructive. I had no idea. That should come in handy.

For the record, the equation above is actually the denominator of a larger equation that I was trying to find the vertical asymptotes for. I know to find those, you just set the denominator to 0, hence the original question. I appreciate all the responses.