Can we solve $nx^2+2x\sqrt{pn^2+q}+rn+s=0$?

Can we solve $nx^2+2x\sqrt{pn^2+q}+rn+s=0$?

Question

Consider the quadratic equation

$$\begin{equation*} nx^2+2x\sqrt{pn^2+q}+rn+s=0, \tag{$*$} \end{equation*}$$

where $p>0$, $p\ne r$ and $n=1$, $2$, $3$, $\dots$.

1. For the case where $p=3$, $q=50$, $r=2$, $s=15$, find the set of values of $n$ for which equation $(star)$ has no real roots.
2. Prove that if $p<r$ and $4q(p-r)>s^2$, then $(star)$ has no real roots for any value of $n$.
3. If $n=1$, $p-r=1$ and $q=s^2/8$, show that $(star)$ has real roots if, and only if, $s\le 4-2\sqrt{2}$ or $s\ge 4+2\sqrt{2}$.

Solution

where $p>0$, $p\ne r$ and $n=1$, $2$, $3$, $\dots$.

1. For the case where $p=3$, $q=50$, $r=2$, $s=15$, find the set of values of $n$ for which equation $(starrepeat)$ has no real roots.

Consider the quadratic equation

$$\begin{equation*} nx^2+2x\sqrt{pn^2+q}+rn+s=0, \tag{$*$} \end{equation*}$$

Let’s first substitute in the values we are given. We want to know for which

$n$

the equation

$$\begin{equation} nx^2+2x\sqrt{3n^2+50}+2n+15=0 \end{equation}$$

has no real roots.

We therefore want to determine the values of

$n$

for which the

is negative, that is,

$n$

such that

$$\begin{align*} &4(3n^2+50)-4n(2n+15) <0\\ \iff\quad & (3n^2+50)-n(2n+15) <0\\ \iff\quad & n^2 -15n +50 <0\\ \iff\quad & (n-5)(n-10) <0.\\ \end{align*}$$

We can now sketch the graph of $y = (n-5)(n-10)$.

We can therefore see that $n^2-15n+50<0$ when $5<n<10$.

Since $n$ is a positive integer, we deduce that the equation $(1)$ has no real roots when $n=6$, $7$, $8$ or $9$.

1. Prove that if $p < r$ and $4q(p-r) > s^2$, then $(starrepeat)$ has no real roots for any value of $n$.

In order for $(starrepeat)$ to have no real roots for a particular value of $n$, its discriminant must be negative for that value of $n$.

In order for $(starrepeat)$ to have no real roots for any value of $n$, its discriminant must therefore be negative for all values of $n$.

The discriminant of

$(starrepeat)$

is

$$4(pn^2+q)-4n(rn+s)=4(p-r)n^2-4sn+4q,$$

so we want to show that

$$4(p-r)n^2-4sn+4q<0$$

for all values of

$n$

, or equivalently,

$$\begin{equation} (p-r)n^2-sn+q<0 \end{equation}$$

for all $n$. But this is a quadratic inequality in $n$, and we want to show that it holds for all values of $n$. (Actually, we only need to show that it is true for positive integer values of $n$. If, though, we can show that it is true for all real values of $n$, this will be stronger than we need and so we will be done.)

First note that $p<r$ (which we are given) means that the leading coefficient of this quadratic is negative, so we know this quadratic is “vertex-up”:

A general quadratic where the coefficient of $x^2$ is negative

In order for this quadratic to be negative for all real $n$, it must have no real roots.

The discriminant of the quadratic in $(2)$ is $s^2-4(p-r)q,$ which is negative, as we are told in the question. So in this case, $(starrepeat)$ has no real roots for any value of $n$.

1. If $n=1$, $p-r=1$ and $q=s^2/8$, show that $(starrepeat)$ has real roots if, and only if, $s\le 4-2\sqrt{2}$ or $s\ge 4+2\sqrt{2}$.

From part (ii), we know that the discriminant of

$(starrepeat)$

is

$$4(p-r)n^2-4sn+4q,$$

which when we’ve substituted in the values given in the question takes the form

$$4-4s+4\frac{s^2}{8},$$

and we want to find the range of values for which this is greater than or equal to

$0$

. So we solve

$$\begin{align*} \frac{s^2}{2}-4s+4&\ge 0 \\ \iff\quad s^2-8s+8&\ge0 \\ \end{align*}$$

We find the roots of this quadratic by completing the square:

$$\begin{align*} &s^2-8s+8=0 \\ \iff\quad &(s-4)^2-16+8=0 \\ \iff\quad &(s-4)^2=8 \\ \iff\quad &s-4=\pm2\sqrt{2}\\ \iff\quad &s=4\pm 2\sqrt{2}. \end{align*}$$

(Alternatively, we could have used the quadratic formula for this.)

Sketching the graph of $y=s^2-8s+8$, we see that it is non-negative if and only if $s\ge 4+2\sqrt{2} \text{ or } s\le 4-2\sqrt{2}.$