Question

If a+b is a root of X square + ax + b equals 0 then what is the maximum value of b square?

Answer 1

If a+b is a root of x^2+ax+b=0, then

$$(a+b)^2+a(a+b)+b=0$$

Hence,

$$a^2+2ab+b^2+a^2+ab+b=0$$

which yields to

$$b^2+b(3a+1)+2a^2=0$$

Therefore,

$$a^2+6a+1\geq 0$$

and

$$b_{1}=\frac{-(3a+1)+\sqrt{a^2+6a+1}}{2},$$ $$b_{2}=\frac{-(3a+1)-\sqrt{a^2+6a+1}}{2}$$

We have

$$a\in(-\infty,-3-2\sqrt{2})\cup(-3+2\sqrt{2},+\infty)$$

If

$$a\in(-\infty,-3-2\sqrt{2}),$$

then b_{1}^{2}>b_{2}^{2}. Hence, in this case b_{1}^{2} is the maximum value of b^2. Otherwise, if

$$a\in(-3+2\sqrt{2},+\infty),$$

then b_{2}^{2}>b_{1}^{2}. In this case, b_{2}^{2} is the maximum value of b^2.