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Homework Help: Need Some Math Help

  1. Jan 1, 2005 #1
    I am doing a pretrig/calc review guide and it is asking me to find all the solutions for each equation.

    -3x=x^2-4

    that is the first equation

    y^4-2y^2=-y^3

    is another one

    any help on what I am supposed to do...I have completly forgotten and have left my book elsewhere
     
  2. jcsd
  3. Jan 1, 2005 #2
    For the first problem add 3x to both sides so you end up with x^2+3x-4=0 with that use the quadratic formula [-b+/-(b^2-4ac)^(1/2)]/2a

    For the second one I would try moving it all to one side and factoring if you can.
     
  4. Jan 1, 2005 #3

    dextercioby

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    The second has an immediate factorization;
    [tex] y^{4}+y^{3}-2y^{2}=0\Rightarrow y^{2}(y^{2}+y-2)=0 [/tex]
    ,which has a double solution y=0 and the other 2 are found by solving the quadratic
    [tex] y^{2}+y-2=0 [/tex]

    Daniel.

    PS.And the first eq.factors as well:(x-3)(x-1)=0.
     
  5. Jan 1, 2005 #4
    Factoring hurts my brain...
     
  6. Jan 1, 2005 #5
    ok thats what I thought ....but I am unsure with

    2x^5=32x^3

    and then the log stuff like

    2logx=log25 I think this one is (2log5)???

    or

    lnx+ln(x+2)=4
     
  7. Jan 1, 2005 #6
    2x^3(16-x^2)=0
    2x^3(4+x)(4-x)=0 difference of squares
    x={0,0,0,4,-4}

    I think.
     
  8. Jan 2, 2005 #7

    dextercioby

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    [tex] 2 \lg x =\lg25\Rightarrow 2\lg x=2\lg5\Rightarrow x=5 [/tex]
    [tex] \ln x+ \ln(x+2) =4\Rightarrow \ln[x(x+2)]=4\Rightarrow x(x+2)=e^{4} [/tex]
    ,which is a second order algebraic equation which can be solved.

    Daniel.
     
    Last edited: Jan 2, 2005
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