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^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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