Minimum value of $a^2 + b^2$ so that the quadratic $x^2 + ax + (b+2) = 0$ has real roots

Question

The equation $ x^2 + ax + (b+2) = 0 $ has real roots, where $a$ and $b$ are real numbers.

How would I find the minimum value of $a^2 + b^2$ ?

Answer 1

Using what you already did ( the discriminant is no-negative and thus $\;a^2-4b\ge8\;$) then

$$a^2+b^2\ge8+4b+b^2=(b+2)^2+4$$

So the above rightmost expression is at least...

Answer 2

In the $(a,b)$ plane, the point $(a,b)$ has to be under the parabola $a^2-4b =8$. The point of this domain closest to origin is the vertex of the parabola, $(0,-2)$. Hence the minimum of $a^2+b^2$ is $4$.

Answer 3

We get the solutions $$ 0 = x^2 + a x + b + 2 = (x + a/2)^2 - a^2/4 + b + 2 \iff \\ x = \frac{-a \pm \sqrt{a^2 - 4b - 8}}{2} $$ which are real for $$ a^2 - 4b - 8 \ge 0 $$ We now want to solve the optimization problem \begin{matrix} \min & a^2 + b^2 \\ \text{w.r.t} & a^2 - 4b - 8 \ge 0 \end{matrix}

The above image shows the function $$ z = f(x,y) = x^2 + y^2 $$ (red surface). It is rotational symmetric around the $z$-axis and has circles in the $x$-$y$-plane with center $(0,0)$ and radius $\sqrt{c}$ as isolines $f(x,y) = c$ (yellow). The curve $$ x^2 - 4y - 8 = 0 \iff \\ y = (1/4) x^2 - 2 $$ (green) is the border of the feasible area $x^2 - 4y -8 \ge 0$, shown in light blue. Point $A=(2,0)$ (pink) is not feasible with the condition, $B=(2,-2)$ (blue) is a feasible point.

Going from inner to outer isolines, we see that the isoline with radius $r=2$, thus $x^2 + y^2 = 2^2 = 4 = c$ is the first one to be part of the feasible area. Thus the minimum is $4$.

Answer 4

If you put b= -2 then then equation becomes an equation with x value as zero which is real..

So minimum value of expression will be four if we put a equals zero..