# Quadratic eq., solve for x when first variable is not 1

• Nov 6th 2009, 01:08 PM
Liselle
Quadratic 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 PM
Raoh
hello(Happy)
$\displaystyle 2x^2-3x-5 =(x+1)(2x-5)$
• Nov 6th 2009, 01:21 PM
Wilmer
Quote:

Originally Posted by Liselle
$\displaystyle 2x^2 - 3x - 5 = 0$
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.

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 PM
Bacterius
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:
Quadratic formula (I won't put the details, they are found on Wiki) :

Say : $\displaystyle ax^2 + bx + c = 0$

You can define a "discriminant" (to know where it comes from, wikipedia) : $\displaystyle \triangle = b^2 - 4ac$

If this discriminant is negative, then the equation has no solution in R and cannot be factorized (you'll see later it has solutions, though).
If this discriminant is zero, then the equation has one root (solution). In fact, it has two, but they are the same. This solution is :

$\displaystyle x = \frac{\ {-b}}{2a}$

If this discriminant is positive, then the equation has two distinct roots :

$\displaystyle x = \frac{\ {-b} {\pm} \sqrt{\triangle}}{2a}$

There you go :)

• Nov 6th 2009, 02:27 PM
westlondongaurav
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 PM
e^(i*pi)
For this equation factoring isn't too difficult but because I'm lazy I nearly always use the quadratic formula