# How do you find the roots, real and imaginary, of y= x^2 - 12x-1  using the quadratic formula?

Sep 30, 2017

See a solution process below:

#### Explanation:

To find the roots we solve ${x}^{2} - 12 x - 1$ for $0$ using the quadratic equation. The quadratic formula states:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 12}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 1}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{\left(- 12\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 12\right)}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{- 1}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{12 \pm \sqrt{144 - \left(- 4\right)}}{2}$

$x = \frac{12 \pm \sqrt{144 + 4}}{2}$

$x = \frac{12 \pm \sqrt{148}}{2}$

$x = \frac{12 - \sqrt{4 \cdot 37}}{2}$ and $x = \frac{12 + \sqrt{4 \cdot 37}}{2}$

$x = \frac{12 - \sqrt{4} \sqrt{37}}{2}$ and $x = \frac{12 + \sqrt{4} \sqrt{37}}{2}$

$x = \frac{12 - 2 \sqrt{37}}{2}$ and $x = \frac{12 + 2 \sqrt{37}}{2}$

$x = \frac{12}{2} - \frac{2 \sqrt{37}}{2}$ and $x = \frac{12}{2} + \frac{2 \sqrt{37}}{2}$

$x = 6 - \sqrt{37}$ and $x = 6 + \sqrt{37}$

Sep 30, 2017

The solutions are $S = \left\{6 + \sqrt{37} , 6 - \sqrt{37}\right\}$

#### Explanation:

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

Our equation is

${x}^{2} - 12 x - 1 = 0$

Start by calculating the discriminant

$\Delta = {b}^{2} - 4 a c = {\left(- 12\right)}^{2} - 4 \cdot \left(1\right) \cdot \left(- 1\right) = 144 + 4 = 148$

As,

$\Delta = 148 > 0$, there are $2$ real roots

The roots are

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- \left(- 12\right) \pm \sqrt{148}}{2 \cdot 1} = \frac{12 \pm \sqrt{148}}{2}$

$\sqrt{148} = 2 \sqrt{37}$

So,

${x}_{1} = 6 + \sqrt{37} = 6 + 6.083 = 12.083$

${x}_{2} = 6 - \sqrt{37} = 6 - 6.083 = - 0.083$

Sep 30, 2017

This quadratic has zeros $x = 6 \pm \sqrt{37}$

#### Explanation:

Note that:

${x}^{2} - 12 x - 1$

is in the standard form

$a {x}^{2} + b x + c$

with $a = 1$, $b = - 12$ and $c = - 1$

This has zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

color(white)(x) = (-(color(blue)(-12))+-sqrt((color(blue)(-12))^2-4(color(blue)(1))(color(blue)(-1))))/(2(color(blue)(1))

$\textcolor{w h i t e}{x} = \frac{12 \pm \sqrt{144 + 4}}{2}$

$\textcolor{w h i t e}{x} = \frac{12 \pm \sqrt{148}}{2}$

$\textcolor{w h i t e}{x} = \frac{12 \pm \sqrt{{2}^{2} \cdot 37}}{2}$

$\textcolor{w h i t e}{x} = \frac{12 \pm 2 \sqrt{37}}{2}$

$\textcolor{w h i t e}{x} = 6 \pm \sqrt{37}$

Bonus

Since $b = - 12$ is large compared to $c = - 1$, we can find good rational approximations to $6 + \sqrt{37}$ rapidly using an integer sequence method.

Define:

$\left\{\begin{matrix}{a}_{0} = 0 \\ {a}_{1} = 1 \\ {a}_{n + 2} = 12 {a}_{n + 1} + {a}_{n} \text{ for } n \ge 0\end{matrix}\right.$

Note that the recurrence rule is chosen such that the sequence:

$1 , x , {x}^{2}$

satisfies it if and only if $x = 6 \pm \sqrt{37}$

Since $\left\mid 6 - \sqrt{37} \right\mid < 1$ the ratio between consecutive pairs of terms will tend to the other, stable, solution $6 + \sqrt{37}$.

The first few terms of this sequence are:

$0 , 1 , 12 , 145 , 1752 , 21169 , 255780$

Then:

$6 + \sqrt{37} \approx \frac{255780}{21169} \approx 12.082762530$

which is correct to $11$ significant digits.

Incidentally, since $c = - a$, we also find:

$6 - \sqrt{37} = - \frac{1}{6 + \sqrt{37}} \approx - \frac{21169}{255780} \approx - 0.08276253030$

which is correct to $10$ significant digits.