The equation $2x^2+ax+(b+3)=0$ has real roots. Find the minimum value of $a^2+b^2$

Question

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?

Answer 1

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.

Answer 2

The minimum value for $ a^2 + b^2 = 8(b+3) + b^2 $, because $ a^2 \geqq 8(b+3) $.

The value of $b$ that minimizes the quadratic $ 8(b+3) + b^2 $ is:

$ 8 + 2b =0 \Rightarrow b = -4$

From $ a^2 \geqq 8(b+3) \Rightarrow b \geqq -3$

Therefore, the minimum value of $ a^2 + b^2 $ is when $ b = -3 $ and $ a = 0 $

$ \Rightarrow a^2 + b^2 \geqq (0)^2 + (-3)^2 $