Question

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

Answer 1

`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`

Answer 2

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`.