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Solve for x: 2x-x^2=3

  1. Aug 28, 2011 #1
    1. The problem statement, all variables and given/known data

    solve for x: 2x-x^2=3

    2. Relevant equations



    3. The attempt at a solution

    2x-x^2=3

    do i go -x^2=3-2x

    thats not right i can tell but i got a warning for not showing work and for this question again i dont know where to start but maybe if i throw a guess in this will satisfy the guy who watches for this.

    actually this is too much of a hassle since im doing so many questions in such a short time period i will be back in another week when i start my actual math course sorry mark im just stressed over doing 8 hours of math everyday.
     
  2. jcsd
  3. Aug 28, 2011 #2
    make the equation = 0
    then either use the quadratic formula or factorising to solve for x
     
  4. Aug 28, 2011 #3

    Mark44

    Staff: Mentor

    Move all of the terms over to one side, so that zero is on the other side. Starting from your first equation,

    2x-x^2=3

    rewrite this as
    -x^2 + 2x - 3 = 0

    You can make the x^2 coefficient positive by multiplying both sides by -1.
     
  5. Aug 28, 2011 #4
    okay I think I will just learn the quad formula for this so with making the x^2 positive would I make the 2x negative and the -3 positive.
     
  6. Aug 28, 2011 #5
    I end up with a -8 in a square root when I use the quad formula what would I do from there or did I screw up somewhere.
     
  7. Aug 28, 2011 #6
    yep there should be a negative number in the square root, which means there are no x solutions

    ie. the graph of the function does not cross the x axis.

    so no real solutions of course there are 'imaginery' solutions
     
  8. Aug 28, 2011 #7
    could I use the quad. formula for a question like 0=125+27k^3
     
  9. Aug 28, 2011 #8
    no the quardratic formula is for something to the power of 2 not 3 or more
    there is a forumula for the power of 3 but it's quite complicated. It is easier to factor out something to end up with a power of 2.
    in this case you can factor out 3x+5
     
  10. Aug 28, 2011 #9
    for 0= 125+27k^3 would the answer be k=5/3
     
  11. Aug 28, 2011 #10
    The formula for factoring the sum of cubes (a^3+b^3) is (a+b)(a^2-2ab+b^2)
    so, (5+3k)(25-30k+9k^2)=0
    so one solution is found by 5+3k=0
    and another by 25-30k+9k^2=0
     
  12. Aug 29, 2011 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The simplest way to solve [itex]125- 27k^3= 0[/itex] is to write it as [itex]27k^3= 125[/itex], divide both sides by 27, and take the cube root of each side. As AracnNoir points out, there are two other solutions (the roots of his [itex]25- 30k+ 9k^2= 0[/tex] but they are both complex numbers.

    Yes, [itex]125= 5^3[/itex] and [itex]27= 3^3[/itex] so 5/3 is an answer. You could have checked that yourself- if x= 5/3, then [itex]x^3= 5^3/3^3= 125/27[/itex] so 27x^3= 27(125/27)= 125.
     
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