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Math Help - Summer review questions!

  1. #1
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    Summer review questions!

    Alright, this summer I have forgotten a lot of what I have learned last year, so I have come to you guys for some help!

    I have a few questions, but I'll probably have to add more later.
    I want to learn how to do these, not just given the answer, could you break it down step by step?

    Here you have to factor and solve for x

    4x^2-3 = 0

    Here, you have to solve for Z:

    4x + 10yz = 0

    y^2+3yz-8z-4x=0

    Thank you
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  2. #2
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    Re: Summer review questions!

    1)
    a^2-b^2 = (a-b)(a+b)
    let a = 2x and b = root(3)
    4x^2 - 3 = [(2x-root(3)][2x+root(3)]

    [(2x-root(3)][2x+root(3)] = 0 means either 2x-root(3) = 0 or 2x+root(3)=0
    so, either x = root(3)/2 or -root(3)/2

    2)
    4x + 10yz = 0 (subtract 4x from both sides)
    10yz = -4x (divide both sides by 10y)
    z = -4x/10y (simplify)
    z = -2x/5y

    3)
    y^2+3yz-8z-4x = 0 (subtract y^2 and add 4x to both sides)
    3yz - 8z = -y^2 + 4x (factor out z)
    z(3y - 8) = -y^2 + 4x (divide by 3y-8)
    z = (-y^2 + 4x)/(3y - 8)
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  3. #3
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    Re: Summer review questions!

    Thank you! I'm just having a bit of a problem on the first one, could you go over it?
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  4. #4
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    Re: Summer review questions!

    Quote Originally Posted by Lotr8808 View Post
    Thank you! I'm just having a bit of a problem on the first one, could you go over it?
    Number one is an example of a difference of squares (a^2 - b^2)
    Anytime a problem looks like this, there is a way to factor it. You need to choose a and b so that a^2 and b^2 fit your problem.

    Since your problem is: 4x^2 - 3 = 0, we need a^2 to be 4x^2 and b^2 to be 3.
    So, a=2x (since (2x)^2 = 4x^2) and b=root(3) (since root(3)^2 = 3)

    Once we choose a and b, we know that it factors to (a-b)(a+b). You can confirm this by foiling.. a^2 + ab - ba - b^2 = a^2 - b^2

    So, the problem factors to (2x-root(3))(2x+root(3)).

    If you have a product of quantities that equals zero, the zero product property states that one or both must be zero. So, to find both solutions of

    [(2x-root(3)][2x+root(3)] = 0

    We set each term in brackets equal to zero and solve for x.

    1) 2x-root(3) = 0 --> 2x = root(3) --> x = root(3)/2
    2) 2x+root(3) = 0 --> 2x = -root(3) --> x = -root(3)/2

    Hope this makes sense.
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  5. #5
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    Re: Summer review questions!

    Great, that really help me, thanks. If I have any more questions, I'll post them here.
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  6. #6
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    Re: Summer review questions!

    Would:
    ln(e^7) = x simplify to 7=x?
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  7. #7
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    Re: Summer review questions!

    Yes, it would.
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