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Please help me about roots

  1. Feb 26, 2008 #1
    if o,p,q are roots of the equation ax^3+bx^2+cx+d=0, determine the equation whose roots are o^2,p^2 and q^2

    who can help me solve it? thank you
     
  2. jcsd
  3. Feb 26, 2008 #2

    malawi_glenn

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    This is missplaced.

    Please study the rules of this forum before posting.
     
  4. Feb 26, 2008 #3

    berkeman

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    Thread moved to Homework Help, Precalc Math.
     
  5. Feb 27, 2008 #4
    Unless I am missing some completely obvious substitution (and I may be), this question is solvable, but not trivial to solve or explain. I'm assuming you want a new function that only has o2, p2, and q2 as roots; thus it will be cubic (barring imaginary roots). Only the constants a, b, c, and d will be different, and we want to write the new constants in terms of a, b, c, and d.

    What you will need to do is to realize that [itex]ax^3+bx^2+cx+d[/itex] and [itex]x^3+(b/a)x^2+(c/a)x + (d/a)[/itex] have the same roots, because the second one is just the first one divided by [itex]a \neq 0[/itex]. The right hand side is still 0, so the solutions aren't changed.

    Now, we know we can factor and get

    [tex]x^3 + \frac{b}{a}x^2 + \frac{c}{a}x + \frac{d}{a} = (x-o)(x-p)(x-q)[/tex]

    Multiply out and find b/a, c/a, and d/a in terms of o, p, and q. Then, suppose that the new cubic equation we want, with roots o2, p2, and q2 is

    [tex]x^3 + ex^2 + fx + g = (x-o^2)(x-p^2)(x-q^2)[/tex]

    Find e, f, and g in terms of o2, p2, and q2. Try to relate it to b/a, c/a, and d/a.

    That last part takes some thought, but give it a try and post back if you're still stuck after trying awhile.
     
    Last edited: Feb 27, 2008
  6. Feb 27, 2008 #5

    rock.freak667

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    Usually in Further math, if we have an equation,[itex]ax^3+bx^2+cx+d=0[/itex] whose roots are [itex]\alpha, \beta, \gamma[/itex] and we want to find an equation whose roots are [itex]\alpha^2,\beta^2, \gamma^2[/itex]. A substitution such as [itex]y=x^2[/itex] gives you it directly.
     
  7. Feb 27, 2008 #6
    Except what happens if the roots are negative? That's the problem I had with substituting y = x2. You will end up with a new equation, but the square root is always positive, so you may not come up with the same solutions. Maybe I am getting myself confused; your idea seems the more reasonable answer.
     
  8. Feb 27, 2008 #7

    rock.freak667

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    sub [itex]x=\sqrt{y}[/itex] into the equation and simplify. It doesn't matter whether or not the roots are -ve or +ve.
     
  9. Feb 27, 2008 #8
    Suppose f(x) = (x + 1)3. The triple root is -1. If we substitute [itex]\sqrt{y}[/itex] we find [itex]f(y) = (\sqrt{y} + 1)^3[/itex] which doesn't have any real roots. Of course, this is only because we define the square root as positive, but I think that is a real concern.
     
  10. Feb 28, 2008 #9

    rock.freak667

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    [itex]f(x)=(x+1)^3[/itex] has root [itex]\alpha (=-1)[/itex] for his equation, there are 3 not one.
     
  11. Feb 29, 2008 #10

    HallsofIvy

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    If a polynomial, with leading coefficient a, has roots o, p, q, then it can be written as a(x-p)(x-q)(x-o)= ax3-a(p+q+o)x2+a (op+ oq+ pq)x- aopq= 0

    Similarly, the equation whose roots are p2, q2, and o2 must be of the form x3- (p2+ q2+ o2)x2- (o2p2+ o2q2+ p2q2)x- o2p2q2= 0.

    You know -aopq= d so opq= d/a and o2p2q2= d2/a2. That's the constant term of the equation you are seeking. In fact if you multiply out (o+ p+ q)2 that will equal b2/a2, that should give you most of what you want.
     
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