Solve 2^2-5X+2=0

I forgot how to do this. I get x=4,1 but that's wrong. I thought you multiply 2*2 and then you know that you need a factor of 4 that adds up to 5 which are 4 and 1.

Then what?

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- Sep 8th 2011, 12:21 PMdanthegreatSolve 2^2-5X+2=0
Solve 2^2-5X+2=0

I forgot how to do this. I get x=4,1 but that's wrong. I thought you multiply 2*2 and then you know that you need a factor of 4 that adds up to 5 which are 4 and 1.

Then what? - Sep 8th 2011, 12:35 PMjnavaRe: Solve 2^2-5X+2=0
$\displaystyle is\ it\ 2^2 - 5x + 2 = 0 , if\ it\ is\, then\ 4 - 5x + 2 = 0 \Rightarrow -5x + 6 = 0 \Rightarrow -5x = -6 \Rightarrow x = \frac{6}{5}.$

- Sep 8th 2011, 12:41 PMskeeterRe: Solve 2^2-5X+2=0
- Sep 8th 2011, 12:46 PMdanthegreatRe: Solve 2^2-5X+2=0
- Sep 8th 2011, 01:10 PMjnavaRe: Solve 2^2-5X+2=0
We know that if this equation is factorable, that we will get the following form (Ax $\displaystyle \pm$ B)(Cx$\displaystyle \pm$D) = 0

He first took $\displaystyle 2x^2$ and factored it. 2 is a prime number so, it's factor's are just 2 and 1, so we know what A and C should be.

(2x $\displaystyle \pm$ B)(1x$\displaystyle \pm$D) = 0

Using FOIL, we know that the last two terms will be multiplied together to come up with a constant. In our case, the constant is 2. Again it is a prime. So B and D are going to be a combination of 1 and 2. Since the last constant in the original equation is a positive number, we know that when multiplied together using FOIL B and D will be both positive or both negative.

Hope that helps some - Sep 8th 2011, 01:55 PMskeeterRe: Solve 2^2-5X+2=0
go to the link and learn ...

Solving Quadratic Equations: Solving by Factoring