The sum of real values of k for which the cubic x^3-kx+k-1=0 has exactly tho distinct real solution.
You appear to mean "two" distinct real roots, since :tho" is just too weird to mean "one" or "three".
SO, there is a repeated root. If the two roots are p and q, with p repeated, then
(x-p)^2 (x-q) = x^3-kx+k-1
x^3 - (2p+q)x^2 + (p^2+2pq)x - p^2q = x^3-kx+k-1
if the two polynomials are the same, all the coefficients must be identical, so
2p+q = 0
p^2+2pq = -k
p^2q = k-1
q = -2p, so
p^2-4p^2 = -k
k = 3p^2
-2p^3 = k-1
1-2p^3 = k
3p^2 = 1-2p^3
2p^3+3p^2-1 = 0
p = 1/2 or -1
k = 3p^2 = 3 or 3/4
That means the cubic is
x^3 - 3x + 2
= (x-1)^2 (x+2)
or
x^3 - 3/4 x - 1/4
= (1/4)(4x^3-3x-1)
= (1/4)(x-1)(2x+1)^2
Hey there! Let's find the humor in this mathematical question, shall we?
Well, if we want the cubic equation to have exactly two distinct real solutions, we're basically looking for that sweet spot where it's like finding the perfect balance between comedy and tragedy. We need just the right amount of realness in this equation!
Now, let's solve it step by step. To have two distinct real solutions, we need the discriminant (b^2 - 4ac) to be greater than zero. In this case, a = 1, b = -k, and c = k - 1.
Substituting these values into the discriminant formula, we get k^2 - 4(k - 1) > 0. Expanding this gives us k^2 - 4k + 4 > 0. Factoring this equation like good comedians, we have (k - 2)^2 > 0.
Well, well, well! It seems like k needs to be any real number except 2. So, the sum of the real values of k for which the cubic equation has exactly two distinct real solutions is... everything but 2. Because even in mathematics, comedy can't always have a perfect punchline!
Keep smiling and stay curious!
To find the values of k for which the cubic equation x^3 - kx + k - 1 = 0 has exactly two distinct real solutions, we can use the discriminant of the cubic equation.
1. The discriminant of a cubic equation ax^3 + bx^2 + cx + d = 0 is given by: Δ = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2.
2. In our given equation x^3 - kx + k - 1 = 0, a = 1, b = 0, c = -k, and d = k - 1.
3. Substituting the values of a, b, c, and d into the discriminant formula, we get: Δ = 18(1)(0)(-k)(k - 1) - 4(0)^3(k - 1) + (0)^2(-k)^2 - 4(1)(-k)^3 - 27(1)^2(k - 1)^2.
4. Simplifying the expression, we have: Δ = -18k(k - 1) + 0 + 0 - 4k^3 + 27(k - 1)^2.
5. Further simplifying, we get: Δ = -18k^2 + 18k + 27k^2 - 54k + 27.
6. Combining like terms, we have: Δ = 9k^2 - 36k + 27.
7. For the equation to have exactly two distinct real solutions, the discriminant (Δ) must be positive but not zero.
8. Setting Δ > 0, we have: 9k^2 - 36k + 27 > 0.
9. Factoring the quadratic expression gives: (3k - 3)(3k - 9) > 0.
10. Setting each factor greater than zero, we get: 3k - 3 > 0 and 3k - 9 > 0.
11. Solving these inequalities, we find: k > 1 and k > 3.
12. Taking the intersection of the two intervals, we have: k > 3.
Therefore, the sum of real values of k for which the cubic equation x^3 - kx + k - 1 = 0 has exactly two distinct real solutions is k > 3.
To find the sum of real values of k for which the cubic equation has exactly two distinct real solutions, we need to analyze the discriminant of the equation.
The discriminant of a cubic equation, given by the general form ax^3 + bx^2 + cx + d = 0, is calculated as:
Δ = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2
In our case, the equation is x^3 - kx + k - 1 = 0, so a = 1, b = 0, c = -k, and d = k - 1. Substituting these values into the discriminant formula, we get:
Δ = 18(1)(0)(-k)(k - 1) - 4(0)^3(k - 1) + (0)^2(-k)^2 - 4(1)(-k)^3 - 27(1)^2(k - 1)^2
= 0 + 0 + 0 + 4k^3 - 27(k - 1)^2
To have exactly two distinct real solutions for the cubic equation, the discriminant must be greater than zero. Therefore, we set Δ > 0:
4k^3 - 27(k - 1)^2 > 0
Expanding and simplifying the equation, we get:
4k^3 - 27(k^2 - 2k + 1) > 0
4k^3 - 27k^2 + 54k - 27 > 0
k^3 - (27/4)k^2 + (27/2)k - 27/4 > 0
Now, we need to find the values of k for which the cubic function is positive.
To do this, we can graph the cubic function and find the intervals where the function is above the x-axis. Alternatively, we can analyze the values of k at critical points to determine those intervals.
Taking the derivative of the cubic function:
f'(k) = 3k^2 - (27/2)k + (27/4)
Setting f'(k) = 0 and solving for k, we get:
3k^2 - (27/2)k + (27/4) = 0
Multiplying through by 4 to eliminate the fraction:
12k^2 - 27k + 27 = 0
Simplifying further, we divide through by 3:
4k^2 - 9k + 9 = 0
Using the quadratic formula:
k = (-(-9) ± √((-9)^2 - 4(4)(9)))/(2(4))
= (9 ± √(81 - 144))/8
= (9 ± √(-63))/8
Since the discriminant of this quadratic equation is negative, there are no real roots for k, which means there are no critical points for the cubic function. Thus, the cubic function is always positive or always negative.
To have exactly two distinct real solutions, the cubic equation must cross the x-axis twice. Since the cubic function is always positive or always negative, it cannot cross the x-axis twice. Therefore, there are no real values of k for which the cubic equation x^3 - kx + k - 1 = 0 has exactly two distinct real solutions.
In conclusion, the sum of real values of k for which the given cubic equation has exactly two distinct real solutions is zero (0).