Given that the equation x^2(3-x) has 3 real solutions of k, give the set of possible values for k.
no equation
no k
no way to help
x^2(3-x)=k. That's the equation.
Well, we know that
y=x^2(3-x)=0 has roots at 0,0,3
And, we know that the graph just touches the x-axis at (0,0) the double root, so cannot have a small k<0 because that will have only a single root near x=3.
So, x^2(3-x) = k only has 3 roots if k>=0
Now, if you have calculus as a tool, you can see that y' = 3x(2-x) so there is a maximum at (2,4). So, we must have k<=4.
So, y=x^2(3-k) has 3 roots only if 0<=k<=4
Ok, thank you.
To find the set of possible values for k, we first need to solve the equation x^2(3-x) = k.
Let's simplify the equation step by step:
1. Distribute x^2:
x^2 * 3 - x^3 = k
2. Rearrange to have all terms on one side:
x^3 + x^2 * 3 - k = 0
3. Now, we know that the equation has 3 real solutions for k. This means that the equation has three distinct roots. In other words, there are three values of x that satisfy the equation for any given value of k.
To determine the set of possible values for k, we need to find the values of k that would make this equation have three distinct real solutions.
One way to determine this is to analyze the nature of the roots of the equation. We can see that the equation is a cubic equation, and the number of real solutions it has can vary based on the discriminant and the position of the roots on the x-axis.
For a cubic equation, the discriminant is given by:
D = (18abc - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2)
In our equation x^3 + x^2 * 3 - k = 0, the coefficients are:
a = 1
b = 3
c = 0
d = -k
Substituting these values into the discriminant formula, we get:
D = (18 * 1 * 3 * 0 - 4 * 3^3 * -k + 3^2 * 0^2 - 4 * 1 * 0^3 - 27 * 1^2 * -k^2)
Simplifying further:
D = (0 - 4 * 27 * -k + 0 - 0 - 27 * k^2)
D = (108k + 27k^2)
For the equation to have three distinct real solutions, the discriminant (D) must be greater than zero.
So, we have:
108k + 27k^2 > 0
To solve this quadratic inequality, we can factor it:
27k(4 + k) > 0
From the equation, we can see that 27k and (4 + k) must have the same signs for the inequality to be true.
Case 1: 27k > 0 and 4 + k > 0
If k > 0 and -4 < k, then both expressions are positive, and the inequality holds.
Case 2: 27k < 0 and 4 + k < 0
If k < 0 and -4 > k, then both expressions are negative, and the inequality holds.
So, the set of possible values for k is:
k ∈ (-∞, -4) ∪ (0, ∞).
Therefore, any value of k less than -4 or greater than 0 would result in the equation x^2(3-x) = k having three real solutions.