Given 2 roots of 2x^3-kx^2+27=0 are equal, find the value of k.

Given 2 roots of 2x^3-kx^2+27=0 are equal, find the value of k.

Working would be much appreciated.

k = 9, the double root is 3 and the third root is -3/2.

A double root means that f(x) = 2x^3 - kx^2 + 27 has the x-axis as a tangent
at the point of the double root. To find this point we need to find where f'(x) = 0.

f'(x) = 6x^2 - 2kx = 0 at either x = 0, (which is clearly not a root), or

6x = 2k ---> x = k/3. Now at this value of x we know that f(x) must equal 0,

so f(k/3) = 2(k/3)^3 - k(k/3)^2 + 27 = (2/27)k^3 - (1/27)k^3 + 27 = 0 --->

-(1/27)k^3 = -27 ---> k^3 = 27^2 = (3^3)^2 = (3^2)^3 ---> k = 9.

So x = k/3 = 9/3 = 3 is the double root, and thus (x - 3)^2 = x^2 - 6x + 9 is

a factor of f(x). After doing the long division I get that

f(x) = (x - 3)^2 * (2x + 3), so x = -3/2 is the third root.

If f(x) is a polynomial with a double root at x = a, then
x = a is also a root of f '(x) = 0.

f(x) = 2x^3 - kx^2 + 27
f '(x) = 6x^2 - 2kx

f '(x) = 0
6x^2 - 2kx = 0
2x(3x - k) = 0
x = 0 or x = k/3

f(0) = 27 so x = 0 is not the double root.
So x = k/3 is the double root.

f(k/3) = 0
2(k/3)^3 - k (k/3)^2 + 27 = 0
(2/27)k^3 - (1/9)k^3 + 27 = 0
(-1/27)k^3 = -27
{take the cube root of both sides}
(-1/3)k = -3
k = 9

CHECK
f(x) = 2x^3 - kx^2 + 27
f(x) = 2x^3 - 9x^2 + 27
f(x) =(x -3)^2 (2x + 3)

NOTES:
If x = a is a double root of some polynomial f(x), then
f(x) = (x - a)^2 g(x) for some polynomial g(x).
f '(x) = 2(x - a) g(x) + (x - a)^2 g'(x)
f '(x) = (x - a)(2g(x) + (x - a)g'(x))

Note f '(a) = 0 so x = a is also a root of f '(x).

Note that this also agrees with Brian's claim that f '(a) = 0.

2x^3-kx^2+27=0

since the degree of this eqn is 3, there should be 3 roits; not 2. what about the 3rd root?


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could it be 2x^2-kx+27=0 ?

if so, then


2x^2 -kx^ - 27=0

roots are equal, so d=b^2-4ac=0

k^2 +216=0



k^2 = -216

k = sqrt [-216]

k = i 6 sqrt 6


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ps Brion is right. sorry about my mistakes.