<s>[INST] 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?
 [/INST] 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 $
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