$\displaystyle 2x^2 - 3x - 5 = 0$

I'm lost as to how I get rid of the 2 in $\displaystyle 2x^2$ so that it's $\displaystyle x^2$. I keep getting fractions that have very high LCDs.

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- Nov 6th 2009, 01:08 PMLiselleQuadratic eq., solve for x when first variable is not 1
$\displaystyle 2x^2 - 3x - 5 = 0$

I'm lost as to how I get rid of the 2 in $\displaystyle 2x^2$ so that it's $\displaystyle x^2$. I keep getting fractions that have very high LCDs. - Nov 6th 2009, 01:17 PMRaoh
hello(Happy)

$\displaystyle 2x^2-3x-5 =(x+1)(2x-5)$ - Nov 6th 2009, 01:21 PMWilmer
Who told you you had to get rid of the 2 ?

2x^2 - 3x - 5 = (2x - 5)(x + 1)

Go here to learn: algebra.help -- Factoring A Trinomial - Nov 6th 2009, 01:28 PMBacteriusReply
You cannot get rid of it (unless dividing everything by $\displaystyle 2$ but then it gets tricky). But you can still factorize this equation :

$\displaystyle 2x^2 - 3x - 5 = (x + 1)(2x - 5)$

But I agree, it is not always obvious to factorize such equations. Have you learnt the quadratic formula yet ? It always works and even tells you when no factorization is possible instead of letting you struggle with an unsolvable problem in R. Here is a spoiler, if you want to know about it :

__Spoiler__: - Nov 6th 2009, 02:27 PMwestlondongaurav
you cannot get rid of the 2, unless you divided the whole equation by 2 which would make the quadratic more difficult. simply factorise the equation:

2x2-3x-5=0

(2x-5)(x+1)=0 - Nov 6th 2009, 02:59 PMe^(i*pi)
For this equation factoring isn't too difficult but because I'm lazy I nearly always use the quadratic formula