Post a New Question

Math (Quadratics)

posted by .

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.

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question