# Roots of polynomials

## Introduction

(I've written this topic specifically for students taking MEI FP1.)

Observe that any quadratic equation can be written in the following form

$$ (x-\alpha)(x-\beta) = 0 $$

Where $\alpha$ and $\beta$ are the quadratic's roots.

Multiply out the left hand side and you get

$$ x^{2} - (\alpha+\beta)x+\alpha\beta = 0 $$

Now observe the general quadratic equation with real coefficients $a$, $b$, and $c$

$$ ax^{2}+bx+c = 0$$

Dividing through by $a$ you get

$$ x^{2} + \frac{b}{a}x+\frac{c}{a} = 0$$

Comparing coefficients with the equation formed by the roots

$$
\begin{align}
\alpha+\beta &= -\frac{b}{a} \\
\alpha\beta &= \frac{c}{a}
\end{align}
$$

We can see that there is a clear relationship between the coefficients of a quadratic and its roots

$$ x^{2} -(\textrm{sum of roots})x + (\textrm{product of roots}) = 0 $$

### Example

Q) Given that $\alpha$ and $\beta$ are the roots of $x^{2}+x-12$, form a new quadratic with roots $\alpha+\frac{1}{\beta}$ and $\beta+\frac{1}{\alpha}$.

A) From the given quadratic we see that

$$
\begin{align}
\alpha+\beta &= -1 \\
\alpha\beta &= -12
\end{align}
$$

So by summing the new roots we get

$$
\begin{align}
\alpha+\frac{1}{\beta}+\beta+\frac{1}{\alpha} &= \alpha+\beta+\frac{\alpha+\beta}{\alpha\beta} \\
&= -1 + \frac{-1}{-12} \\
&= -\frac{11}{12}
\end{align}
$$

And by multiplying the new roots together we get

$$
\begin{align}
\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right) &= \alpha\beta+2+\frac{1}{\alpha\beta} \\
&= -12 +2 + \frac{1}{-12} \\
&= -\frac{121}{12}
\end{align}
$$

Therefore the new quadratic is $12x^{2}+11x-121$.

### Example

Q) Given that $\alpha$ and $\beta$ are the roots of $10x^{2}+x-3$, form a new quadratic with roots $2\alpha-1$ and $2\beta-1$.

A) From the given quadratic we see that

$$
\begin{align}
\alpha+\beta &= -\frac{1}{10} \\
\alpha\beta &= -\frac{3}{10}
\end{align}
$$

So by summing the new roots we get

$$
\begin{align}
2\alpha-1+2\beta-1 &= 2(\alpha+\beta)-2 \\
&= 2\left(-\frac{1}{10}\right)-2 \\
&= -\frac{11}{5}
\end{align}
$$

And by multiplying the new roots together we get

$$
\begin{align}
(2\alpha -1)(2\beta -1) &= 4\alpha\beta-2(\alpha+\beta)+1 \\
&= 4\left(-\frac{3}{10} \right)-2\left(-\frac{1}{10} \right)+1 \\
&= 0
\end{align}
$$

Therefore the new quadratic is $5x^{2}+11x$.