The equation $2x^2+ax+(b+3)=0$ has real roots. Find the minimum value of $a^2+b^2$ Question:
The equation $2x^2+ax+(b+3)=0$ has real roots. Find the minimum value of $a^2+b^2$
My working:
$D=a^2-8(b+3)\geqq0$
$a^2\geqq8(b+3)=8b+24$
Add $b^2$ to both sides
$a^2+b^2\geqq b^2+8b+24=(b+4)^2+8\geqq 8$
$\therefore {(a^2+b^2)
}_{min} = 8$
However the answer in the book is 9.
What have I done wrong?

By your work: $$a^2+b^2=a^2+b^2-\frac{3}{4}(a^2-8(b+3))+\frac{3}{4}(a^2-8(b+3))\geq$$
$$\geq a^2+b^2-\frac{3}{4}(a^2-8(b+3))=\frac{1}{4}(a^2+4(b+3)^2)+9\geq9.$$
The equality occurs for $a=0$ and $b=-3$, which says that we got a minimal value.
