Thread: 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?

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 ...



if so ...



,

Originally Posted by skeeter

did you really mean ...



if so ...



,

where did you get 1 and 2? (bolded)

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 B)(Cx D) = 0

He first took 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 B)(1x 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

Originally Posted by danthegreat

where did you get 1 and 2? (bolded)

go to the link and learn ...

Solving Quadratic Equations: Solving by Factoring