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

##### 3 Answers

See a solution process below:

#### Explanation:

To find the roots we solve

For

Substituting:

The solutions are

#### Explanation:

The quadratic equation is

Our equation is

Start by calculating the discriminant

As,

The roots are

So,

This quadratic has zeros

#### Explanation:

Note that:

#x^2-12x-1#

is in the standard form

#ax^2+bx+c#

with

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

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

Since

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

Incidentally, since

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

which is correct to