# Math Help - Solve 2^2-5X+2=0

1. ## Solve 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?

2. ## Re: Solve 2^2-5X+2=0

$is\ it\ 2^2 - 5x + 2 = 0 , if\ it\ is\, then\ 4 - 5x + 2 = 0 \Rightarrow -5x + 6 = 0 \Rightarrow -5x = -6 \Rightarrow x = \frac{6}{5}.$

3. ## Re: Solve 2^2-5X+2=0

Originally Posted by danthegreat
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?
did you really mean ...

$2x^2 - 5x + 2 = 0$

if so ...

$(2x - 1)(x - 2) = 0$

$x = \frac{1}{2}$ , $x = 2$

4. ## Re: Solve 2^2-5X+2=0

Originally Posted by skeeter
did you really mean ...

$2x^2 - 5x + 2 = 0$

if so ...

$(2x - 1)(x - 2) = 0$

$x = \frac{1}{2}$ , $x = 2$
where did you get 1 and 2? (bolded)

5. ## Re: Solve 2^2-5X+2=0

Originally Posted by danthegreat
where did you get 1 and 2? (bolded)
We know that if this equation is factorable, that we will get the following form (Ax $\pm$ B)(Cx $\pm$D) = 0

He first took $2x^2$ 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 $\pm$ B)(1x $\pm$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

6. ## Re: Solve 2^2-5X+2=0

Originally Posted by danthegreat
where did you get 1 and 2? (bolded)
go to the link and learn ...

Solving Quadratic Equations: Solving by Factoring