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Math Help - root and equation

  1. #1
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    root and equation

    if a and b are the roots of the quadratic equationb ax^2 +bx +c =0 ,express (alpha-2beta)(2alpha-beta) in term if a,b,c . deduce the condition that one root of the equation is twice the other roots.

    answer provided :2b^2 = 9ac, (2b^2-9ac)/a^2



    2)if,a,b,c is subset of real number, with a is not equal to zero, and the roots of the quadratic equation ax^2+bx+c = 0 are real , show that the roots of ay^2 - (b^2-2ac)y + c^2 = 0 are also real. if the roots of quadratic equation ax^2 + bx +c = 0 states the value of alpha + beta and alpha Xbeta in terms of a,b,c . hence find the roots of the second equation in terms of a and b.

    answer provided -b/a,c/a, alpha^2,beta^2
    Last edited by qweiop90; July 29th 2008 at 08:58 AM.
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  2. #2
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    1) You can't be serious. If a and b are roots of that quadratic, then necessarily a^{3}+ab+c = 0 and b^{2}(a+1) + c = 0.

    2) Are you SURE you mean to use 'a' and 'b' in both places? Perhaps you mean to say that A and B are roots of ax^{2}+bx+c=0?
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    1) You can't be serious. If a and b are roots of that quadratic, then necessarily a^{3}+ab+c = 0 and b^{2}(a+1) + c = 0.

    2) Are you SURE you mean to use 'a' and 'b' in both places? Perhaps you mean to say that A and B are roots of ax^{2}+bx+c=0?
    . i cant be serious? that is the question i copied frm the books?
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  4. #4
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    Why did you change to "alpha" and "beta" later?

    We have \alpha\;=\;\frac{-b+\sqrt{b^{2}-4ac}}{2a} and \beta\;=\;\frac{-b-\sqrt{b^{2}-4ac}}{2a}.

    Can you impose the conditions of the problem statement on those two roots of the equation?
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  5. #5
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    oops, i thought you already know . sorry then.

    dont get what you mean? currently, i can proved that 2b^2= 9ac. but i couldnt get 1st answer
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  6. #6
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    Well, \beta has to be the lesser of the two, so \alpha = 2\beta. Substitute the larger expressions and see where it leads.
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  7. #7
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    Quote Originally Posted by TKHunny View Post
    Well, \beta has to be the lesser of the two, so \alpha = 2\beta. Substitute the larger expressions and see where it leads.
    alpha = 2 beta?i get what you mean,but can you state the starting, and for the further steps, i will try my hardest effort to solve.

    for the question2, which method i should applied with?
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  8. #8
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    \alpha\;=\;2*\beta\;\implies\;\frac{-b+\sqrt{b^{2}-4ac}}{2a}\;=2*\frac{-b-\sqrt{b^{2}-4ac}}{2a}

    2) Just use the Quadratic Formula (or Complete the Square if you wish.) It does not matter what the coefficients are, just as long as it LOOKS like a quadratic equation.

    Example: If (Frog)x^2 + (Horse)x + Butterfly = 0, then x = \frac{-Horse \pm \sqrt{Horse^{2}-4*Frog*Butterfly}}{2*Frog}
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