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Thread: College Algebra

  1. #11
    Elite Member mmm4444bot's Avatar
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    Oops. I never clicked "Submit". Oh well, I'll post it anyway.

    Yes, I'll look at the rest of your exercises later.

    Quote Originally Posted by twhitley

    no I'm not adding them , I remember learning to factor and if it added up to the 2nd number then it was right

    Oh, okay, now I understand what you're talking about. That four is the coefficient on the x-term in the polynomial.

    Yes, you are adding them, heh, heh.

    You are correct. When factoring a quadratic polynomial into a product of binomials, you're looking for two numbers that add up to the coefficient on the x-term.


    Solving for a, I think of it as solving for x or y, is this correct?

    Sure. In the equation a^2 + 4a - 45 = 0, we can think of the variable a as x and 0 as y.
    To solve for the equation a^2 + 4a = 45 means to find all Real numbers that when squared and added to four of themselves end up being 45.

    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  2. #12
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    Re: College Algebra

    Quote Originally Posted by twhitley
    2) Let f(x)= x^2+2x-2 and g(x)= 2-3x. Find the following:
    a) f(-1)
    b) g(-3)
    c) f(a+b)



    If anyone can help, it would be greatly appreciated!
    You're given

    f(x) = x[sup:3choz3ku]2[/sup:3choz3ku] + 2x - 2

    If you want to know what f(3) is, then you'd substitute 3 for each x:

    f(x) = x[sup:3choz3ku]2[/sup:3choz3ku] + 2x - 2
    f(3) = 3[sup:3choz3ku]2[/sup:3choz3ku] + 2(3) - 2

    Then, do the arithmetic on the right side.....
    f(3) = (3)(3) + 2(3) - 2
    f(3) = 9 + 6 - 2
    f(3) = 15 - 2
    f(3) = 13

    So, f(3) = 13

    Try that approach on your problems.

  3. #13
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    Re: College Algebra

    cool, that even made it that much more clear! Thanks so much !

  4. #14
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    Re: College Algebra

    so then f(-1)= (-1)(-1)+2(-1)-2
    = 1+-2-2
    = -3

    so f(-1) = -3

    Is this correct?

  5. #15
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    Re: College Algebra

    Quote Originally Posted by twhitley
    4) sove this system of equations:
    5x+4y=-8
    2x-y=-11
    Multiply the 2nd equation by 4: 8x - 4y = -44

    Then add them:
    5x + 4y = -8
    8x - 4y = -44
    ===========
    13x = -52

    Can you finish off?
    I'm just an imagination of your figment !

  6. #16
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    Re: College Algebra

    so this would mean x=-4? also, why do I multiply the second equation by 4?

  7. #17
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    Re: College Algebra

    then if I plugged the x value in the equation (-4) then I would get y=3 so would my final answer be x=-4 and y=3

  8. #18
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    Re: College Algebra

    I think I would multiply the second equation by 4 so the y would cancel out so I can find value of x, is this correct

  9. #19
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by twhitley

    f(-1) = -3

    Is this correct? Yes
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  10. #20
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    Quote Originally Posted by twhitley

    3)solve for x: (2/x-1) +1=(2/x^2-x)
    If a ratio equals 1, then the numerator and denominator must be the same.

    Subtract 2/(x - 1) from both sides.

    1 = 2/(x^2 - x) - 2/(x - 1)

    Do you know how to combine the two ratios on the righthand side ?

    Get a common denominator. Multiply the second ratio by x/x, then both denominators will be the same.

    1 = 2/(x^2 - x) - 2x/(x^2 - x)

    1 = (2 - 2x)/(x^2 - x)

    Since the top and bottom of the righthand side must be the same number, set the numerator equal to the denominator, and solve for x.

    Make sure that you check both answers!

    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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