Recent content by Erfan1

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    MHB Roots of polynomial equations ( Substitution )

    How do I reduce u^4 + 5u^3 + 6u^2 + 5u + 1 = 0 to v^2 + 5v + 4 = 0 by using v = u + 1/u ?
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    MHB Mean value of second derivative of a function

    The curve C is defined parametrically by x = 4t - t^2 and y = 1 - e^-t where 0≤t<2 . Show that the mean value of d^2 y / dx^2 with respect to x over the interval 0≤x≤7/4 is (4e^(-1/2) - 3)/ 21 . I've figured out d^2 y/dx^2 as ((t-1)e^-t)/(4(2-t)^3) . Any idea how to do the the other part ?
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    MHB Roots of polynomial equations 4

    The product of two of the roots of the equation ax^4 + bx^3 + cx^2 + dx + e = 0 is equal to the product of the other two roots. Prove that a*d^2 = b^2 * e
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    MHB Roots of polynomial equations 3

    Obtain the sum of the squares of the roots of the equation x^4 + 3x^3 + 5x^2 + 12x + 4 = 0 . Deduce that this equation does not have more than 2 real roots . Show that , in fact , the equation has exactly 2 real roots in the interval -3 < x < 0 . Denoting these roots α and β , and the other...
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    MHB Roots of polynomial equations 2

    The roots of the equation x^3 - x - 1 = 0 are α β γ and S(n) = α^n + β^n + γ^n (i) Use the relation y = x^2 to show that α^2, β^2 ,γ^2 are the roots of the equation y^3 - 2y^2 + y - 1 =0 (ii) Hence, or otherwise , find the value of S(4) . (iii) Find the values of S(8) , S(12) and S(16)I have...
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    MHB Roots of polynomial equations 1

    Re: Roots of polynomial equations I'm not really good at complex number .
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    MHB Roots of polynomial equations 1

    Find the sum of the squares of the roots of the equation x^3 + x + 12 = 0 and deduce that only one of the roots is real . The real root of the equation is denoted by alpha . Prove that -3< alpha < -2 , and hence prove that the modulus of each of the other roots lies between 2 and root 6 . I...
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    MHB Summation of Series: Prove 1/4 - 1/(2n+2)

    Partial fractions ! I tried this 1/(r*(r+1)*(r+2)) = 1/(2r) - 1/(r+1) + 1/(2(r+2)) but it seems that this doesn't work :D
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    MHB Summation of Series: Prove 1/4 - 1/(2n+2)

    Prove that 1/(1*2*3) + 1/(2*3*4) + ... + 1/(n*(n+1)*(n+2)) = 1/4 - 1/(2*(n+1)*(n+2) .
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