High school quadratic inequality question

Question

Any help with the following question would be very much appreciated.

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

Please note that, unfortunately, the question provides no clarification regarding the nature of $a$ and/or $b$.

Answer 1

The standard treatment of this type of question is by C-S inequality but with the functional flavor. Start out by rewriting the equation as:

$(2x^2+3)^2 = (-ax-b)^2 = (ax+b)^2\le (1+x^2)(a^2+b^2)\implies a^2+b^2 \ge \dfrac{(2y+3)^2}{1+y}$, with $y = x^2 \ge 0$. Put $t = y+1 \ge 1 \implies a^2+b^2 \ge \dfrac{(2t+1)^2}{t}=f(t)$. We have: $f'(t) = 4-\dfrac{1}{t^2} > 0$ on $[1,\infty)\implies f_{\text{min}}(t) = f(1) = 9 \implies (a^2+b^2)_{\text{min}} = 9$ as claimed.

Answer 2

Observe that $$x=\frac{-a \pm \sqrt{a^2-8b-24}}{4}.$$ For real $a$ and $b$, $x$ is real if and only if $a^2-8b-24 \ge 0$. Equivalently, $$a^2 \ge 8b+24 \implies a^2 +b^2 \ge b^2 + 8b+24 =(b+4)^2+8.$$ The minimum value is achieved when $$a^2=8b+24 \ge 0 \implies b \ge -3$$ Further, for $b \ge -3$, $(b+4)^2+8$ achieves its minimum when $b = -3$. Therefore, the minimum value of $a^2+b^2$ is $(b+4)^2+8 = \color{red}{9}$ which occurs when $\color{red}{b=-3}$ and $\color{red}{a=0}$.

Note that, when $\color{red}{b=-3}$ and $\color{red}{a=0}$, both roots are zero.

Answer 3

This is an exercise in recognizing how you can use the knowledge you have to break a problem into smaller steps.

First, solve this exercise:

What are the conditions on $a$ and $b$ for which $2x^2 + ax + (b + 3) = 0$ has real roots?

which is probably going to be substituting the appropriate formulas into a theorem you know about when quadratic equations have real roots (although other approaches are possible). Then, you solve the second exercise:

$a$ and $b$ satisfy some conditions. Find the minimum value for $a^2 + b^2$.

Where, of course, you replace "some conditions" with your answer to the first exercise.

Answer 4

For the First part you can either use the quadratic equation or completing the square. From there you should have what Moo commented as a hint.

The next thing you will need to do in order to make sure you have real roots is make sure that the value under the $\sqrt{}$ is non-negative. so $a^2-8(b+3) \ge 0$

From here we can get

$a^2 \ge 8b+24 \ge 0 \implies b\ge -3 $.

$a^2 \ge 8b+24 \implies a^2+b^2 \ge b^2+8b+24$.
note that $b^2 +8b+24 = (b+4)^2+8$

Since we know $b \ge -3$ we can see that the minimum for $(b+4)^2+8$ must happen at $b=-3$ and $a=0$

$ (\color{blue}{-3}+4)^2+8=9 $

$ (\color{blue}{-2}+4)^2+8=12 $

Answer 5

$b=-2x^2-ax-3$.

Thus, by AM-GM and C-S $$a^2+b^2=a^2+(2x^2+ax+3)^2=$$ $$=(1+x^2)a^2+2x(2x^2+3)a+(2x^2+3)^2=$$ $$=(1+x^2)a^2+4x^2(x^2+3)+2x(2x^2+3)a+9\geq$$

$$\geq2|ax|\sqrt{(x^2+1)(4x^2+12)}+2x(2x^2+3)a+9\geq$$ $$\geq2|ax|\sqrt{(x^2+1)(4x^2+9)}+2x(2x^2+3)a+9\geq$$ $$\geq2|ax|(2x^2+3)+2x(2x^2+3)a+9\geq9.$$ The equality occurs for $a=x=0$, which gives that $9$ is a minimal value.

Done!