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Homework Help: Quadratic equation with roots of α and β

  1. May 15, 2010 #1
    1. The problem statement, all variables and given/known data

    The quadratric equation x^2 - 5x + 7 = 0 has the roots α and β. Find the quadratic equation with roots α^3 and β^3.

    2. Relevant equations


    3. The attempt at a solution

    u = α^3
    α = u^(1/3)

    u^(2/3) - 5(1/3) + 7 = 0

    u^(1/3)[u - 5] = -7 --- minus 7 off each side followed by taking cubic root u out of the left side.

    u[u - 5] = -343 --- put everything to the power of 3.

    u^2 - 5u + 343 = 0

    but the answer is u^2 – 20u + 343 = 0

    I cannot see where I've gone wrong - help me please :) thank you
  2. jcsd
  3. May 15, 2010 #2


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    Well, for starters, when you cubed both sides of the equation, you forgot to cube the (u-5) term...
  4. May 15, 2010 #3


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    Surely you mean u= x^3 and x= u^(1/3)

    but you didn't. You should have u[u- 5]^3= -343- and multiplying that out gives a 4th degree equation not a quadratic equation.

    where you've gone wrong is in not doing what you thought you did! You did not cube the entire equation.

    You should try, instead, [itex](x- \alpha^3)(x- \beta^3)= x^2- (\alpha^3+ \beta^3)x+ \alpha^3\beta^3[/itex].

    knowing that [itex]\alpha[/itex] and [itex]\beta[/itex] satisfy [itex]x^2- 5x+ 7= 0[/itex]] tells you that [itex]\alpha+ \beta= -5[/itex] and [itex]\alpha\beta= 7[/itex].

    That makes it easy to see that [itex]\alpha^3\beta^3= (\alpha\beta)^3= 7^3[/itex].

    For [itex]\alpha^3+\beta^3[/itex] use the fact that [itex]\alpha^3+ \beta^3= (\alpha+ \beta)(\alpha^2+ \alpha\beta+ \beta^2)[/itex].

    Note that since [itex]\alpha+ \beta= -5[/itex], [itex](\alpha+ \beta)^2= \alpha^2+ 2\alpha\beta+ \beta^2= \alpha^2+ 2(7)+ \beta^2= 25[/itex] so that [itex]\alpha^2+ \beta^2= 11[/itex].
    Last edited by a moderator: May 15, 2010
  5. May 15, 2010 #4
    Thank you for the replies guys - I'll have another go :)
  6. May 15, 2010 #5


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    or another way of doing it:



    on cubing both sides, [tex]u(u-15u^{2/3}+75u^{1/3}-125)=-343[/tex]

    But we know from the first equation that [tex]u^{2/3}-5u^{1/3}+7=0[/tex] so then [tex]-15(u^{2/3}-5u^{1/3}+7)=-15u^{2/3}+75u^{1/3}-105=0[/tex]

    Subbing this into our result, [tex]u(u-20)=-343[/tex] hence [tex]u^2-20u+343=0[/tex] as required.
  7. May 15, 2010 #6
    Ahh - this is what I was attempting to do but failed :P

    Thank you for the input Mentallic :)
  8. May 15, 2010 #7


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    I had a feeling that that was the approach you were looking for :wink:
  9. May 15, 2010 #8
    I see when I was taking u^(1/3) out I only left u inside instead of u^(1/3) which messed the rest of the answer up :P
  10. May 15, 2010 #9
    Sorry - how did you get the u(u-20) part. Slightly confused by that part :S
  11. May 15, 2010 #10


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    Keep this in mind.

    Now after cubing we have,


    But we want to get rid of the fractional powers, and notice how the [tex]u^{2/3}[/tex] term has a multiplier of -15, let's multiply the first equation by -15.

    So looking at,




    Substitute this into the other equality,


    What do you end up with?
  12. May 15, 2010 #11
    Sorry- still not getting it :|

    from [tex]u(u-15u^{2/3}+75u^{1/3}-125)=-343[/tex] to.. [tex]u(u - 20)[/tex]
  13. May 15, 2010 #12


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    Ok how about this,

    let [tex]y=u^{2/3}-5u^{1/3}+7[/tex]

    therefore [tex]-15y=-15u^{2/3}+75u^{1/3}-105[/tex]

    What we have is,




    which is equivalent to,




    Notice that part in there that is equal to -15y. Let's substitute that into it then,


    Do you get this?

    But the only difference is that we had y=0, so you get the answer that you need.

    If you don't understand the process, don't use it. HallsofIvy's method is perfectly fine too.
  14. May 15, 2010 #13
    I fully understand now - thank you very much :D
  15. May 19, 2010 #14
    [itex]\alpha^3+ \beta^3= (\alpha+ \beta)(\alpha^2 - \alpha\beta+ \beta^2)[/itex].
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