Question

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

Answer 1

See a solution process below:

Explanation:

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

For `color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0`, the values of `x` which are the solutions to the equation are given by:

`x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))`

Substituting:

`color(red)(1)` for `color(red)(a)`

`color(blue)(-12)` for `color(blue)(b)`

`color(green)(-1)` for `color(green)(c)` gives:

`x = (-color(blue)((-12)) +- sqrt(color(blue)((-12))^2 - (4 * color(red)(1) * color(green)(-1))))/(2 * color(red)(1))`

`x = (12 +- sqrt(144 - (-4)))/2`

`x = (12 +- sqrt(144 + 4))/2`

`x = (12 +- sqrt(148))/2`

`x = (12 - sqrt(4 * 37))/2` and `x = (12 + sqrt(4 * 37))/2`

`x = (12 - sqrt(4)sqrt(37))/2` and `x = (12 + sqrt(4)sqrt(37))/2`

`x = (12 - 2sqrt(37))/2` and `x = (12 + 2sqrt(37))/2`

`x = 12/2 - (2sqrt(37))/2` and `x = 12/2 + (2sqrt(37))/2`

`x = 6 - sqrt(37)` and `x = 6 + sqrt(37)`

Answer 2

The solutions are `S={6+sqrt37, 6-sqrt37}`

Explanation:

The quadratic equation is

`ax^2+bx+c=0`

Our equation is

`x^2-12x-1=0`

Start by calculating the discriminant

`Delta=b^2-4ac=(-12)^2-4*(1)*(-1)=144+4=148`

As,

`Delta=148>0`, there are `2` real roots

The roots are

`x=(-b+-sqrtDelta)/(2a)=(-(-12)+-sqrt(148))/(2*1)=(12+-sqrt148)/2`

`sqrt148=2sqrt37`

So,

`x_1=6+sqrt37=6+6.083=12.083`

`x_2=6-sqrt37=6-6.083=-0.083`

Answer 3

This quadratic has zeros `x=6+-sqrt(37)`

Explanation:

Note that:

`x^2-12x-1`

is in the standard form

`ax^2+bx+c`

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

This has zeros given by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

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

`color(white)(x) = (12+-sqrt(144+4))/2`

`color(white)(x) = (12+-sqrt(148))/2`

`color(white)(x) = (12+-sqrt(2^2*37))/2`

`color(white)(x) = (12+-2sqrt(37))/2`

`color(white)(x) = 6+-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:

`{ (a_0 = 0), (a_1 = 1), (a_(n+2) = 12a_(n+1)+a_n " for " n >= 0) :}`

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

`1, x, x^2`

satisfies it if and only if `x = 6+-sqrt(37)`

Since `abs(6-sqrt(37)) < 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) ~~ 255780/21169 ~~ 12.082762530`

which is correct to `11` significant digits.

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

`6 - sqrt(37) = -1/(6+sqrt(37)) ~~ -21169/255780 ~~ -0.08276253030`

which is correct to `10` significant digits.