1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.


    that is the first equation


    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


    User Avatar
    Science Advisor
    Homework Helper

    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]


    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


    and then the log stuff like

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


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

    I think.
  8. Jan 2, 2005 #7


    User Avatar
    Science Advisor
    Homework Helper

    [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.

    Last edited: Jan 2, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook