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Math Help - Solving Quadradics

  1. #1
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    Exclamation Solving Quadradics

    Hi, I just got done doing a bunch of trig stuff, which I understand fairly well, but my teacher included some algebra review problems with quadratic stuff, etc. and i can't remember how to do these for the life of me. Thank you for any and all help.

    Solve the following equations without a calculator.

    a. (x-7)(x+5) = 20

    b. 2y^2 = y

    c. 4y^2 = 1

    d. 3y^3 + 2y^2 = 5y

    e. 2y - (1/y) = 1
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by leviathanwave View Post
    Hi, I just got done doing a bunch of trig stuff, which I understand fairly well, but my teacher included some algebra review problems with quadratic stuff, etc. and i can't remember how to do these for the life of me. Thank you for any and all help.

    Solve the following equations without a calculator.

    a. (x-7)(x+5) = 20

    b. 2y^2 = y

    c. 4y^2 = 1

    d. 3y^3 + 2y^2 = 5y

    e. 2y - (1/y) = 1
    a. (x-7)(x+5) = 20

    here we first multiply out. the easy way to do this is to chose one term in one of the brackets and multiply everything in the other bracket, then take the other term and multiply everything in the other bracket. so for this, i take x and mulitply each term in the other bracket and then -7 and multiply everything. here goes:

    (x - 7)(x + 5) = 20
    => x^2 + 5x - 7x - 35 = 20
    => x^2 + 5x - 7x - 35 - 20 = 0
    => x^2 - 2x - 55 = 0 ........now it's your standard quadratic problem.
    now we try to find two numbers that when multiplied gives -55 and when added gives -2. no such luck. we need the quadratic formula here then (or completing the square if you know it).

    the quadratic formula says:
    given a quadratic function of the form
    ax^2 + bx + c = 0
    we can find x using the formula

    x = [-b +/- sqrt(b^2 - 4ac)]/2a

    so for x^2 - 2x - 55 = 0, a = 1, b = -2, c = - 55
    so x = [2 +/- sqrt(4 -4(1)(-55)]/2
    => x = [2 +/- sqrt(4 + 220)]/2
    => x = [2 +/- sqrt(224)]/2
    => x = (2 + sqrt(224))/2 or (2 - sqrt(224))/2
    => x = 1 + sqrt(56) or 1 - sqrt(56)



    b. 2y^2 = y .............first get the y's on one side

    => 2y^2 - y = 0 ...............now factor the common y out
    => y(2y - 1) = 0 ..............now, if two functions multiplied give 0, one or the other must be 0
    so y = 0 or 2y - 1 = 0
    => y = 0 or y = 1/2


    c. 4y^2 = 1 ............simply solve for y by transposing the formula, ("transpose" is just a fancy word for moving things about until you have what you want on one side of the equal sign by itself)

    4y^2 = 1

    => y^2 = 1/4 ............divided both sides by 4
    => y = +/- sqrt(1/4)
    => y = +/- 1/2 ..............note, it can be +1/2 or -1/2 since either multipied by itself gives +1/4


    d. 3y^3 + 2y^2 = 5y ............transpose
    => 3y^3 + 2y^2 - 5y = 0 .......now factor out the common y
    => y(3y^2 + 2y - 5) = 0
    => y = 0 or 3y^2 + 2y - 5 = 0

    3y^2 + 2y - 5 = 0
    (3y + 5)(y - 1) = 0 ............if you need help with this step tell me
    => 3y + 5 = 0 or y - 1 = 0
    so all solutions are:

    y = 0, y = -5/3 or y = 1


    e. 2y - (1/y) = 1

    we dont like that 1/y, its ugly and it makes things complicated, let's get rid of it by multiplying through by y

    2y - (1/y) = 1
    => 2y^2 - 1 = y ...........now transpose
    => 2y^2 - y - 1 = 0 .......now foil
    => (2y + 1)(y - 1) = 0 .........again, if you need help with this step, tell me
    => 2y + 1 = 0 or y - 1 = 0
    so y = -1/2 or y = 1
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  3. #3
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    Smile Thanks!

    Thank you very much, your post was very helpful and informative. You cleared a couple of things up for me, i was wondering if i could set separate factors equal to 0 or not. Thanks again for the help ^_^
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