Ask Me Help Desk

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)
-   Online Education (https://www.askmehelpdesk.com/forumdisplay.php?f=401)
-   -   Algebra II problem solvers, with work? (https://www.askmehelpdesk.com/showthread.php?t=488704)

  • Jul 15, 2010, 07:21 PM
    MandyMarieLove
    Algebra II problem solvers, with work?
    Hey, I need help with my Algebra II class. I have found math solver sites, but none give me the steps to do the problem! I don't want to be given the answers and not know what to do! Helppp?
  • Jul 15, 2010, 08:09 PM
    KISS

    Not quite the way to learn.

    Post away with your thoughts and possible solutions.
  • Jul 15, 2010, 08:26 PM
    MandyMarieLove

    Okay, so here's my big thing.. solving quadratic equations by factoring, I know it sounds stupid.. But, I don't understand the factoring process of it. Can you help?

    Here's an example:

    Identify the solutions to the equation 2x2 + 5x = 3 when you solve by factoring.
  • Jul 15, 2010, 08:43 PM
    ISneezeFunny

    solving by factoring:

    step 1: move everything to one side of the equation, leaving 0 on the other side.

    2x^2 + 5x - 3 = 0

    step 2: divide by the number in front of the highest factor (in this case 2), however, it will make this very ugly, so we'll skip that.

    step 3: find factors

    2x^2 + 5x - 3 = 0

    (2x - 1)(x + 3) = 0

    step 4: FOIL it out to see if that works

    2x^2 + 6x - x - 3 = 0

    step 5: solve

    (2x - 1)(x + 3) = 0

    2x - 1 = 0
    x + 3 = 0

    2x = 1
    x = 1/2

    x = -3
  • Jul 15, 2010, 08:50 PM
    KISS

    Assuming the 2x2 is 2x^2

    Start by making it equal to zero

    2x^2+5x-3 = 0

    There are only two things that can get you 2x^2 and that's a 2x and an x

    The only factors of 3 are 3 and 1; one has to be negative to end up with a -3

    try them and arrive at:

    (x+3) (2x-1) = 0 is the answer.

  • All times are GMT -7. The time now is 01:29 PM.