Question

Determine the values of "k" for which the equation x^3-12x+k = 0 will have 3 different roots.

Answer 1

ALL real?

Answer 2

There are a few different approaches to take here. Are you allowed to use calc? Also, have you tried graphing it in geoegbra, making "k" a slider?

Answer 3

no i am not allow to use calculus or calculator

Answer 4

maybe graphing calculator

Answer 5

Veieta is probably one way.

Answer 6

\[x^3-12x+k=(x-a)(x-b)(x-c)\]
\[x^3-12x+k=(x^2-ax-bx+ab)(x-c)\]
\[x^3-12x+k=(x^2+(-a-b)x+ab)(x-c)\]
\[x^3-12x+k=(x^3+(-a-b)x^2+abx-cx^2-c(-a-b)x-abc)\]
\[x^3-12x+k=x^3+(-a-b-c)x^2+(ab+ac+bc)x-abc\]
So we have
-a-b-c=0
ab+ac+bc=-12
k=-abc

Answer 7

Myin just cooked vieta's theorem for you :)

Answer 8

where a, b,c are the roots of course

Answer 9

First assume the equation as x^3 -12x = 0 the graph would be like this:

If you want 3 different roots, that means to say the graph will intersect with the x axis 3 times.

Looking at the equation in the question x^3 -12x +k, k will shift the graph up and down along the y axis. Thus u need to find the range of values of k such that, U want the in-between of 2 situations:

as u can see in the 2 pictures, the curve only cut the x axis 2 times (not what you want).

SO now that you know what u want to find, we can solve the question proper.

First u differentiate the equation.
y= x^3 -12x
dy/dx = 3x^2 - 12
let dy/dx= 0 ( you want to find the maximum and minimum point)
0= 3x^2 -12
x^2 = 4
x= 2 or -2 ( remember (-2)^2 also = 4)

Solve for y.
when x =2
y=x^3 -12x
y= 2^3 -12(2)
y= -16

when x = -2
y= (-2)^3 -12(-2)
y= -8+24
y= 16
thus the max point is (-2,16) and the min point is (2, - 16)
Since you want the curve to intersect the x axis 3 times, you can only shift the curve up and down less than 16 units.

thus -16<k<16