How do you find the roots of #x^3-x^2-17x+15=0#?

2 Answers
Jun 6, 2017

Answer:

#x ~~ 6.0664 or 0.9336270#

Explanation:

#x^3 - x^2 - 17x + 15 = 0#

#x^3 - x^2 = x^2#

#x^2 - 17x + 15 = 0#

Now we can use the quadratic equation to solve.

#ax^2 + bx + c = 0#

#x^2 - 17x + 15 = 0#

#a = 1#
#b= -17#
#c = 15#

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

# x = (--17 +- sqrt(-17^2-4 xx 1 xx 15)) / (2 xx 1) #

We can simplify some things that we don't need to make the equation more simple.

# x = (cancel(--)17 +- sqrt(-17^2-4 cancel(xx 1) xx 15)) / (2 cancel(xx 1)) #

# x = (17 +- sqrt(-17^2-4 xx 15)) / 2 #

# x = (17 +- sqrt(289 -4 xx 15)) / 2 #

# x = (17 +- sqrt(289 -60)) / 2 #

# x = (17 +- sqrt(229)) / 2 #

# x = (17 +- 15.132746) / 2 #

# x_1 = (17 + 15.132746) / 2 #

# x_1 = 32.132746 / 2 #

# color(blue)(x_1 = 16.0664 #

# x_2 = (17 - 15.132746) / 2 #

# x_2 = 1.86725405 / 2 #

# color(blue)(x_2 = 0.9336270 #

Jun 6, 2017

Answer:

Three real roots:

#x_n = 1/3(1+4sqrt(13) cos(1/3 cos^(-1)(-(125sqrt(13))/1352)+(2npi)/3))#

for #n = 0, 1, 2#

Explanation:

Given:

#x^3-x^2-17x+15 = 0#

First use a Tschirnhaus transformation:

#0 = 27(x^3-x^2-17x+15)#

#color(white)(0) = 27x^3-27x^2-459x+405#

#color(white)(0) = (3x-1)^3-156(3x-1)+250#

#color(white)(0) = t^3-156t+250" "# where #" "t=3x-1#

Let:

#k=2sqrt(-(color(blue)(-156))/3) = 4sqrt(13)#

Note that:

#k^3/4 = 52 k = 208sqrt(13)#

Substitute:

#t = k cos theta#

Then our equation becomes:

#0 = (k cos theta)^3-156(k cos theta)+250#

#color(white)(0) = 208sqrt(13)(4 cos^3 theta - 3 cos theta)+250#

#color(white)(0) = 208sqrt(13)cos 3 theta+250#

So:

#cos 3 theta = -250/(208sqrt(13)) = -(125sqrt(13))/1352#

So:

#3 theta = +-cos^(-1)(-(125sqrt(13))/1352)+2npi#

So:

#cos theta = cos(1/3cos^(-1)(-(125sqrt(13))/1352)+(2npi)/3)#

So:

#t_n = 4sqrt(13) cos(1/3cos^(-1)(-(125sqrt(13))/1352)+(2npi)/3)#

with distinct roots for #n = 0, 1, 2#

Hence solutions of the original cubic:

#x_n = 1/3(1+4sqrt(13) cos(1/3 cos^(-1)(-(125sqrt(13))/1352)+(2npi)/3))#

for #n = 0, 1, 2#

#x_0 ~~ 4.19825#

#x_1 ~~ -4.07504#

#x_2 ~~ 0.876781#

graph{x^3-x^2-17x+15 [-10, 10, -52, 52]}

#color(white)()#
Footnote

I suspect there is a typo in the question. If the sign on the #x^2# term was #+# instead of #-#, then we find:

#0 = x^3+x^2-17x+15#

#color(white)(0) = (x-1)(x^2+2x-15)#

#color(white)(0) = (x-1)(x+5)(x-3)#

and hence roots #1#, #-5# and #3#.