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!

Solve for x

  1. Jan 7, 2016 #1
    1. The problem statement, all variables and given/known data
    (x2+3x+3)1/3 + (2x2+3x+2)1/3 = 6x2+12x+8
    2.Relevant equations


    3. The attempt at a solution
    (x2+3x+3)1/3 -1 +(2x2+3x+2)1/3 -1 = 6x2+12x+6
    (x2+3x+2)/((((x2+3x+3)1/3)2 + (x2+3x+3)1/3 +1) + (2x2 +3x+1)/((((2x2+3x+2)1/3)2)+(2x2+3x+2)1/3 +1) -6(x+1)2=0
    then
    x=-1
    or
    (x+2)/((((2x2+3x+2)1/3)2)+(2x2+3x+2)1/3 +1)-6(x+1)=0 (*)
    now there is 1 root left it is also -1 but no common factor in (*)
    sorry i am not good at english
     
  2. jcsd
  3. Jan 7, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hello emrys, :welcome:

    To be honest, I have no idea how to deal with this kind of equation analytically, and I can't follow your first step (*), so I simply plotted the terms and I see that only -1 is a solution.

    Multiplicity is 2, also from visual inspection.

    Is there a question in your post ?

    upload_2016-1-7_12-54-50.png


    (*) perhaps you can explain, then I will learn something from this too :smile: !
     
    Last edited: Jan 7, 2016
  4. Jan 7, 2016 #3
    for
    A= (x2+3x+2)1/3
    B= (2x2+3x+2)1/3
    then thanks to calculator i know that x = -1
    so i substitute x for -1 to find value of A,B then subtract exactly that value to find common factor (x+1):
    A -1 + B -1 = 6x2 +12x+6
    (A-1)(A2+A*1+12)/(A2+A*1+12) + (B - 1)(B2+B*1+12)/(B2+B*1+12)-6(x+1)2=0
    (A3-13)/(A2+A*1+12) + (B3-13)/(B2+B*1+12) -6(x+1)2=0
    then you got
    x=-1
    or
    (x+2)/(A2+A*1+12) + (2x+1)/(B2+B*1+12) - 6(x+1)=0(*)
     
  5. Jan 7, 2016 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    A= (x2+3x+3)1/3

    So basically you are also solving this numerically ? Finding x = -1 to satisfy the original equation ?
    Then what is the purpose of the remainder of the work ?
     
  6. Jan 7, 2016 #5
    solve(*) to find the final root (also = -1).
    to end the problem you must find all the roots or the equation becomes an impossible equation.
    and there is 1 root left in (*).
    In brief, i want you guys to help me solve (*) (normally do it like you didnt know about x=-1)
     
  7. Jan 7, 2016 #6
    I am actually around this level of math. My first approach at this question would be to graph it and get the solutions by treating it like a system of equations, so ##y=(x^2 + 3x + 3)^1/3 + (2x^2 + 3x + 2)^1/3## and ##y = 6x^2 + 12x + 8## (sorry if I did the latex wrong). If you need to do it algebraicly, it still may help having the answer to help guide you in solving it.
     
  8. Jan 7, 2016 #7
    And I did do the latex wrong. It should be to the 1/3 power not to the 1st power decided by 3.
     
  9. Jan 7, 2016 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I have a very strong suspicion that your equation cannot be solved "analytically" in any easy way, so if I were trying to do it the very first thing I would do is plot the functions. That reveals x = -1 as a double root (a fact that can be verified analytically, once we know what we are looking for). That is how I would normally do it if I did not know about x = -1 ahead of time.
     
  10. Jan 9, 2016 #9
    Use cauchy inequality, we can solve this easily.
     
  11. Jan 9, 2016 #10

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    How can you use an inequality to solve an equation? You should at least show how you would do it.
     
  12. Jan 9, 2016 #11

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    @Isaac0427
    Regarding the LaTeX:

    To include the /3 in the the exponent, place { } around the 1/3 like so: ^{1/3} Doing that to your first LaTeX expression gives:

    ##y=(x^2 + 3x + 3)^{1/3} + (2x^2 + 3x + 2)^{1/3}\ ##​
     
  13. Jan 10, 2016 #12
    we will do it again from beginning
    we have
    (x2+3x+3)1/3>0 ∀x∈ℝ
    (2x2+3x+2)1/3>0 ∀x∈ℝ
    and 6x2+12x+8>0 ∀x∈ℝ
    then
    apply cauchy inequality (A+A1+A2+...+An)/n ≥(A*A1*A2*....*An)1/n; (A,A1,A2,...,An≥0)
    6x2+12x+8=(1*1(x2+3x+3))1/3+(1*1(2x2+3x+2))1/3≤(x2+3x+3+1+1)/3+(2x2+3x+2+1+1)/3=x2+2x+3
    ⇔6x2+12x+8≤x2+2x+3
    ⇔52+10x+5≤0
    ⇔5(x+1)2≤0
    ⇔x=-1
     
  14. Jan 10, 2016 #13

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Looks good.

    There is a typo in the indicated line. Of course, it should read:

    5x2+10x+5≤0​
     
  15. Jan 10, 2016 #14
    My bad. it's 5x2
     
  16. Jan 10, 2016 #15

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your method works accidentally in this one special case. In most cases it will not work. For instance, if you change the equation to ##A + B = 6x^2+12x + 6## (where ##A,B## are as before) you will find that ##x^2 + 2 x + 3 \geq 6 x^2 + 12 x + 6## at any solution ##x## of the original equation. This is true, but it does not help you to find ##x##. Now we need to use a numerical method, to find there are two roots: ##x = -1.6116857121542768293## and ##x = -0.37972871671526324113##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Solve for x
  1. Solve for x (Replies: 7)

  2. Solve for x (Replies: 19)

  3. Solve for x (Replies: 6)

  4. Solve for (x) (Replies: 3)

  5. Solve for x (Replies: 3)

Loading...