x^2-x+2x-2

Question

x^2-x+2x-2 hellllllllllllllllllllllllllllllllllllllllllllllppppppppppp

Answer 1

If you want to solve the equation x^2-x+2x-2=0 then first simplify it to x^2-x-2 = 0.

If we have ax^2+ 2bx + c =0 we can solve the equation as follow in two steps:

step one ) calculate Δ = sqrt(b^2 - 4 ac) if Δ≥0 then the equation can be solved. go to step two.

note1: In our case  Δ = sqrt(1+8) = 3 ≥0 then go to note2.

step two) calculate the roots as x1 = (-b + Δ)/2 and x2 = (-b - Δ)/2

note2: x1 = (-(-1) +3)/2 = 1 and x2 = (-(-1) - 3)/2 = -2

Now you not only have the roots of the equation but also can rewrite your equation as

x^2-x+2x-2 = (x-x1) (x-x2) = (x-1)(x+2)

Answer 2

Hi Ashley! Looks like you got a big 'ol algebra expression there. I say expression, because there's no equal signs, so it's not technically an equation.

If you are looking to simplify that expression, I can help.

You can use the distributive property here. Let's separate our expression a little:

x^2 - x    + 2x -2

Do the two terms on the left have anything in common? Yeah, they both have an x!

Let's pull that out. x^2 - x = x (x - 1)

Do the terms on the right have anything in common? Yeah, they both have a 2!

Let's pull that out. 2x - 2 = 2 (x - 1)

Now we have x^2 - x + 2x - 2 = x (x - 1) + 2 (x - 1)

Hmmm. It looks like we have some like terms here, because the terms in parentheses are the same. We can add or subtract them like any other variable.

So we have x (x -1) + 2 (x - 1) = (x + 2) (x - 1)

Does that work? We can use the FOIL method to check ourselves. First Outer Inner Last

Yep, sure does.

Hope this helps, Ashley!