show that -k is a root of x^3 +(1-k^2)x + k,find the other two roots and?

Question

find set of values of k for which f has exactly one real root

Answer 1

1) Show that (-k) is a root of x^3 + (1 - k^2)x + k.

x^3 + (1 - k^2)x + k = 0

x^3 + x - k^2x + k = 0

-(k + x)(kx - x^2 - 1) = 0



=> k + x = 0

x = 0 - k

x = - k (first solution)



=> kx - x^2 - 1 = 0

kx - x^2 = 1

x(k - x) = 1

x = 1 / (k - x)

x = 1 / k - 1 / x



2) Find the other two roots.

From the second solve for x, follows that:



x = 1 / 2 (k - √k^2 - √4)

x = 1 / 2 (k - k - 2)

x = (0 - 2) / 2

x = - 2 / 2

x = - 1 (second solution)



x = 1 / 2 (k + √k^2 - √4)

x = 1 / 2 (k + k - 2)

x = (2k - 2) / 2

x = 2k / 2 - 2 / 2

x = k - 1 (third solution)



The two solutions are shown above.



3) Find set of values (of k) for which f has exactly one real root.

x^3 + (1 - k^2)x + k = 0

x^3 + x - k^2x + k = 0

k - k^2x = -x^3 - x

k = k^2x - x^3 - x

k = -x(x^2 - k^2 + 1)



=> x ≠ 0 and k = x + 1 / x

k = -x



Hope this helped!



~Maths

Answer 2

if x = - k

is a root then

x + k is a factor and the problem could be:

(x + k)(x^2 + ax + b) = x^3 + (1 - k^2)x + k



multiplying the left we have:

x^3 + (a + k)x^2 + (ak + b)x + bk



thus:

a + k = 0

ak + b = 1 - k^2

bk = k



from the third equation,

b = 1



plugging this into the second equation:

ak + 1 = 1 - k^2

ak = - k^2



provided k doesnt equal 0,

a = -k



actually from the coefficient of x^2

we obtain a = - k as well.



so the problem is:

(x + k)(x^2 - kx + 1)



solve:

x^2 - kx + 1 = 0

to find the other two roots:



i will go with completing the square:

(x - k/2)^2 = - 1 + k^2/4

x = k/2 +/- (1/2)sqrt(k^2 - 4)



or

x = (1/2)*[k + sqrt(k^2 - 4)]

and

x = (1/2)*[k - sqrt(k^2 - 4)]

Answer 3

(-k - x) (-1 + k x - x^2)