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Math Help - algebra problem

  1. #1
    Newbie
    Joined
    Jan 2009
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    Unhappy algebra problem

    need ur help people on few questions


    Q1>>>Show that the roots of the equation
    (x a) (x b) + (x b) (x c) + (x c) (x a) = 0
    are always real and they cannot be equal unless a = b = c

    Q2>>If w is a complex cube root of unity, then show that
    (1 + 5w2 + w4) (1 + 5w + w2) (5 + w + w2) =64

    Q3>>>A question paper contains 12 questions divided into two sections. Section I contains
    7 questions and section II contains 5 questions. In how many ways can a candidate
    choose the questions if he has to select 8 questions in all with teh restriction of
    atleast 3 questions from each section?

    Q4>>Let A =
    |1 -1 0 |
    |2 2 4 |
    |2 3 4 |

    and B =

    |4 2 4 |
    |0 1 2 |
    |2 1 5 |
    Find AB.

    Use this to solve the following system of equations:
    x y = 3
    2x + 3y + 4z = 17
    y + 2z = 7


    thanku for ur help frndz
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  2. #2
    MHF Contributor red_dog's Avatar
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    Medgidia, Romania
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    Q1)
    The equation can be written as 3x^2-2(a+b+c)x+ab+ac+bc=0
    The discriminant is \Delta=4(a^2+b^2+c^2-ab-ac-bc)

    But it is well known that a^2+b^2+c^2\geq ab+ac+bc \ \forall a,b,c\in\mathbf{R}

    so, \Delta\geq 0 and the roots are real.

    The equality stands if and only if a=b=c
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  3. #3
    MHF Contributor red_dog's Avatar
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    Medgidia, Romania
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    Q2)

    We have w^3=1 and w^2+w+1=0.

    Then w^4=w^3\cdot w=w

    1+5w^2+w^4=(1+w+w^2)+4w^2=4w^2

    1+5w+w^2=(1+w+w^2)+4w=4w

    5+w+w^2=(1+w+w^2)+4=4

    Then the product is equal to 4w^2\cdot 4w\cdot 4=64w^3=64
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