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Thread: College Algebra

  1. #1
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    College Algebra

    Need Help getting started on the following problems, it's been almost 10 years since I have had any algebra, but it was always my favorite subject, so if I know I'm doing it correctly, I will catch back on.

    1) solve for a: a^2 + 4a=45

    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)

    3)solve for x: (2/x-1) +1=(2/x^2-x)

    4) sove this system of equations:

    5x+4y=-8
    2x-y=-11


    If anyone can help, it would be greatly appreciated!

  2. #2
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    Re: College Algebra

    [tex]a^2+4a \ = \ 45, \ \implies \ a^2+4a-45 \ = \ 0.[/tex]

    [tex]Hence, \ using \ grouping, \ we \ get \ a^2+9a-5a-45 \ = \ 0[/tex]

    [tex]Then \ factoring, \ we \ get \ a(a+9)-5(a+9), \ gives \ (a+9)(a-5) \ = \ 0[/tex]

    [tex]Therefore, \ by \ the \ zero \ property \ rule \ a \ = \ -9 \ or \ a \ = \ 5.[/tex]

    [tex]I'll \ leave \ the \ check \ up \ to \ you.[/tex]
    I am not, therefore I do not think. Contrapositive of Descartes' quip.

  3. #3
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    Re: College Algebra

    I get all of this except after you factor, you said a= -9 or 5 and when I factored I used 9, -5 thus this would = the positive 4, can you explain, thanks.

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    Quote Originally Posted by twhitley

    I get all of this except after you factor, you said a= -9 or 5 ? Glenn is talking about the solutions, here, not the factorization

    and when I factored I used 9, -5 ? This is correct; Glenn also used 9 and -5 in the factorization

    thus this would = the positive 4 Your use of the pronoun "this" is ambiguous.

    You're not adding the two numbers in the factorization together, are you?
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  5. #5
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by twhitley

    1) solve for a: a^2 + 4a=45
    Do you understand the meaning of the phrase "solve for a" ?

    If not, then please tell us right away because that is where we would need to begin helping you.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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    Re: College Algebra

    no I'm not adding them , I remember learning to factor and if it added up to the 2nd number then it was right, with that disregarded.
    Solving for a, I think of it as solving for x or y, is this correct?
    there is an (a) and (a^2) so I guess this is where I'm confused.
    Thanks.

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    Re: College Algebra

    [tex](a+9)(a-5) \ = \ 0[/tex]

    [tex]Now, \ if \ a \ = \ -9 \ or \ if \ a \ = \ 5, \ the \ above \ equation \ rings \ true.[/tex]

    [tex]Check: \ (-9+9)(-9-5) \ = \ 0, \ 0(-14) \ = \ 0, \ 0 \ = \ 0,[/tex]

    [tex]and \ (5+9)(5-5) \ = \ 0, \ (14)(0) \ = \ 0, \ 0 \ = \ 0[/tex]
    I am not, therefore I do not think. Contrapositive of Descartes' quip.

  8. #8
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    Re: College Algebra

    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, with that disregarded.
    Solving for a, I think of it as solving for x or y, is this correct?
    there is an (a) and (a^2) so I guess this is where I'm confused.
    Thanks.
    If the product of two or more factors is 0, then at least one of those factors must be 0.

    Symbolically: if a*b = 0, then a = 0 or b = 0.

    We often apply this principal to solve quadratic (or higher degree) equations which can be factored. For example, if we have to solve

    x[sup:tthh96kl]3[/sup:tthh96kl] - 5x[sup:tthh96kl]2[/sup:tthh96kl] + 6x = 0, we might start by factoring:

    x(x[sup:tthh96kl]2[/sup:tthh96kl] - 5x + 6) = 0
    x(x - 2)(x - 3) = 0

    Now, since we have three factors whose product is 0, we know that at least one of those factors must have a value of 0.

    So....since x(x - 2)(x - 3) = 0, we can say
    x = 0
    OR
    x - 2 = 0, which means x = 2
    OR
    x - 3 = 0, which means x = 3.

    So we have three solutions: x = 0 or x = 2 or x = 3

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    Re: College Algebra

    ok, I get that , does this mean to answer the question I would put both down, a= -9 or a=5, my thought process was to only have one answer for a.
    thanks!

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    Re: College Algebra

    cool , i got it , thank all of you soooo much, have you looked at any of the others?

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