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Algebra problem need help on final step

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data
    2. Relevant equations
    3. The attempt at a solution
    Now I'm not sure what I should do next...
    Maybe divide both sides by 2 to get rid of the squares
    Then say that [itex]ax = 13a[/itex] ?
    I am confused.
  2. jcsd
  3. Jan 28, 2013 #2


    Staff: Mentor

    In your attempt at a solution where did the -5ax term go?

    It seems you just dropped it. Oh I see no you can't add the x^2 and x terms together to get 13ax^2.

    The strategy should be to group terms so that the right hand side of the equation is a quadratic in x and the left hand side is 0.

    then find the roots of the quadratic.
  4. Jan 28, 2013 #3


    Staff: Mentor

    Is there something you're supposed to do with this, like, say, solve for x?
    You have a mistake above. Apparently you added 5ax to 8x2 and got 13ax2. They are not like terms, so can't be combined.
    Dividing by 2 doesn't get rid of exponents.
    There's an error above, as well. You divided the left side by 13, but divided the right side by 2. You can't do that.
  5. Jan 28, 2013 #4
    Ok so I thought about the problem again and now I have



    But now I don't see any like terms or anything I can reduce or eliminate
  6. Jan 28, 2013 #5
    Bring it to the format 1c110885bd9155bea6b6630e7d24d6c4.png and
    just use the quadratic root equation


    in your case it would be [itex]8 x^2 - 26a^2 + 5ax=0[/itex]
    Last edited: Jan 28, 2013
  7. Jan 28, 2013 #6
    I don't understand how to enter my equation into that formula
  8. Jan 28, 2013 #7
    in you equation [tex]x=x[/tex]
    [tex]a= 8[/tex]
    [tex]b= +5a[/tex]
    [tex]c= -26a ^ 2[/tex]
  9. Jan 28, 2013 #8


    Staff: Mentor

    It would be better as 8x2 + 5ax - 26a2 = 0
  10. Jan 29, 2013 #9
    Does it tell you to solve for X, it's a bit confusing, since you have left it unmentioned or I am missing something. Anywho, what you cannot do for certain is divide the sides of the equal sign with x or a. If it were x you were solving for, if you divide it like that, then one of your solutions will disappear, the number of the dimension of the variable tells you how many solutions there will have to be.

    You can treat both of the numbers a and x as variables here. 1 time, solve for x and then solve again for a, since you haven't said what needs to be done. I have a hunch one needs to solve for x here.
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