## how to find real solutions to x^2 + 3x - 4 = 0

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
Germo
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Joined: Fri Feb 27, 2009 12:35 am
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### how to find real solutions to x^2 + 3x - 4 = 0

Easy question, just missed one part of the lesson:

I know that x2 + 3x - 4 = 0 has 2 solutions because of the degree of the equation. So my first assumption was to factor out a x and make the problem x(x+3)+4, but I realized that that would give me 3 solutions. So how exactly do I go about getting the real solution?

DAiv
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Joined: Tue Dec 16, 2008 7:47 pm
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### Re: x^2+3x-4=0

Hi Germo, welcome to the forums.

The lesson on Factoring Quadratics explains how to do this.

DAiv

FrozenBlood
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Joined: Wed Feb 04, 2009 11:31 pm
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### Re: how to find real solutions to x^2 + 3x - 4 = 0

You didn't factor the equation properly.

The equation is this in factored form: (x + 4)(x - 1) = 0. You have to ask yourself, which two numbers whose product is -4 also adds up to 3 (the x-coefficient of the second term)? The answer to that is 4 and -1. And thus, the equation factors as (x + 4)(x - 1) = 0.

Then when you set each of the factors equal to zero, you get the two following solutions: x = {-4, 1}