Posted by Drake on Wednesday, June 26, 2013 at 10:41am.
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)  Steve, Wednesday, June 26, 2013 at 11:25am
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^2ab+b^2) + (a+b) + 6ab
= (a+b)(a^2+b^2+1) + ab(6(a+b))
= 2(a^2+b^2+1) + ab(62)
= 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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