Review question

# Can we solve $nx^2+2x\sqrt{pn^2+q}+rn+s=0$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6246

## Solution

Consider the quadratic equation $\begin{equation*} nx^2+2x\sqrt{pn^2+q}+rn+s=0, \tag{*} \label{eq:starrepeat} \end{equation*}$

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 $\eqref{eq:starrepeat}$ has no real roots.
Let’s first substitute in the values we are given. We want to know for which $n$ the equation $$$nx^2+2x\sqrt{3n^2+50}+2n+15=0 \label{eq:1}$$$

has no real roots.

We therefore want to determine the values of $n$ for which the discriminant 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 $\eqref{eq:1}$ has no real roots when $n=6$, $7$, $8$ or $9$.

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

In order for $\eqref{eq:starrepeat}$ to have no real roots for a particular value of $n$, its discriminant must be negative for that value of $n$.

In order for $\eqref{eq:starrepeat}$ to have no real roots for any value of $n$, its discriminant must therefore be negative for all values of $n$.

The discriminant of $\eqref{eq:starrepeat}$ 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, $$$(p-r)n^2-sn+q<0 \label{eq:2}$$$

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”:

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

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

1. If $n=1$, $p-r=1$ and $q=s^2/8$, show that $\eqref{eq:starrepeat}$ 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 $\eqref{eq:starrepeat}$ 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 $s\ge 4+2\sqrt{2} \text{ or } s\le 4-2\sqrt{2}.$