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

3 Answers
Sep 30, 2017

Answer:

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)#

Sep 30, 2017

Answer:

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#

Sep 30, 2017

Answer:

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.