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Math (Quadratics)

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If the quadratic equation for x
{1+(a+b)^2}x^2−2(2a+2b+1)x+5=0
has real roots, what is the value of
a^3+b^3+6ab+a+b?

  • Math (Quadratics) - ,

    the discriminant must not be negative, so

    4(2a+2b+1)^2 - 4*5(1+(a+b)^2) >= 0
    a^2 + 2ab + b^2 - 4(a+b) + 4 >= 0
    (a+b)^2 - 4(a+b) + 4 >= 0
    ((a+b)-2)^2 >= 0
    Hmmm. That's true for any a,b.

    If a+b=2, then the original equation has a repeated root and

    a^3+b^3+6ab+a+b = (a+b)(a^2-ab+b^2) + (a+b) + 6ab
    = (a+b)(a^2+b^2+1) + ab(6-(a+b))
    = 2(a^2+b^2+1) + ab(6-2)
    = 2a^2+2b^2+2 + 4ab
    = 2(a^2+2ab+b^2)+2
    = 2(a+b)^2 + 2
    = 2*4+2
    = 10

    Haven't yet worked it out for other a,b but the wording of the question makes me believe the expression is 10 for any a,b.

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