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Suppose $r$ and $s$ are the values of $x$ that satisfy the equation $x^2 - 2mx + (m^2+2m+3) = 0$ for some real number $m$. Find the minimum real value of $r^2+s^2$.

My Solution:

Use the quadratic formula:

$ax^2 + bx + c = 0 \implies x = \frac{1}{2}\left(-b \pm \sqrt{b^2 - 4ac}\right)$

($a=1, b=-2m,$ and $c=m^2+2m+3$)

We get $r = m + \sqrt{-(2m + 3)}$ and $s = m - \sqrt{-(2m + 3)}$

Then $r^2 + s^2 = 2\left(m^2 - 2m +3\right)$

This is a concave-up parabola with vertex at $m=1$.

Then $\left(r^2 + s^2\right)\big|_{m = 1} = 2\left(1 - 2 + 3\right) = 4.$

This answer is wrong, but why?

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    What are $r$ and $s$ when $m=1$? Perhaps $r$ and $s$ are supposed to be real numbers? – 2017-02-26

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Since $r$ and $s$ are roots of the equation, we have that $(x-r)(x-s)=0$

Then $x^{2} - (r+s)x +rs=0$

Comparing coefficients:

$r+s = 2m$

$rs = m^{2} + 2m + 3$

$\Rightarrow (r+s)^{2} = 4m^{2}$

$\Rightarrow r^{2}+ s^{2} = 4m^{2} - 2rs = 4m^{2} -2m^{2} -4m -6$

$r^{2}+ s^{2} = 2m^{2} -4m -6$

So we now just need to find the minimum value of this function of $m$:

$r^{2}+ s^{2} = 2[(m-1)^{2} -4]$

Which we can clearly see has a minimum of -$8$ at $m=1$

Edit:

But then we also require the initial equation to have real roots

$\Rightarrow 4m^{2} - 4m^{2} -8m - 12 \geq 0$

$\Rightarrow m \leq -\frac{3}{2}$

It is then simple to see by considering the graph of $m$ that $m= -\frac{3}{2}$ would provide the minimum value of $r^{2} + s^{2} = \frac{9}{2}$

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    This is not the minimum of $r^2+s^2$, it is only the minimum of $2m^2-4m-6$.. – 2017-02-26
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    Thanks for the spot, I have corrected it now. – 2017-02-26
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You don't need to use the quadratic formula; $r^s+s^2$ is a symmetric polynomial in $r$ and $s$, hence it can be expressed as a function of the elementary symmetric functions: $$S=r+s=2m,\quad P=rs=m^2+2m+3.$$ Indeed $\;r^2+s^2=S^2-2P=2m^2-4m-6$. This is a quadratic polynomial in $m$, and its minimum is attained at $\;m=-\dfrac{-4}{2\cdot 2}=1$. However, we must take into account that the given equation has roots if and only if its reduced discriminant $$\Delta'=m^2-(m^2+2m+3)=-2m-3$$ is non--negative, i.e. if $m\in\bigl(-\infty,-3/2\bigr]$. On this interval, $2m^2-4m-6$ is decreasing, so the minimum, knowing there exists real roots, is $$2m^2-4m-6\Big\rvert_{m=-3/2}=\frac92.$$

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solving the quadratic equation we get $$x_1=m+\sqrt{-2m-3}$$ $$x_2=m-\sqrt{-2m-3}$$ then we get $$x_1^2+x_2^2=2m^2-4m-6$$ can you proceed? i have got $$min(2m^2-4m-6)$$ under the condition $$-2m-3\geq 0$$ is equal to $$\frac{9}{2}$$ and $$m_{min}=-\frac{3}{2}$$