kx^2 + (k+2)x -3 =0 has roots which are real and positive. Find the possible values that k may have.
I understand that I have to use the discriminant to solve this, but I can't seem to get the answer right, so I really need help!!
Hello
It's a given that we have two roots.
The roots must be real and distinct at that. So the disc. has to be nonzero and positive.
The two roots must also be positive. This means that both the product and the sum of the roots ought to be positive.
Suppose (alpha) and (beta) are the two roots.I'd construct a sign diagram, k against the disc., as well as k against (alpha)*(beta) and k against (alpha)+(beta),and eventually find the values for k such that the values of the three (disc., (alpha)*(beta) and (alpha)+(beta)) are positive simultaneously.
To find the possible values for k such that the given quadratic equation has real and positive roots, we can start by analyzing the discriminant (denoted as Δ) of the quadratic equation.
The discriminant is calculated as follows: Δ = b^2 - 4ac
where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.
In the given equation, kx^2 + (k+2)x - 3 = 0, we have:
a = k, b = k+2, and c = -3.
Now, let's plug these values into the discriminant formula:
Δ = (k+2)^2 - 4(k)(-3)
= k^2 + 4k + 4 - 12k
= k^2 - 8k + 4
Since we want the roots to be real and positive, the discriminant (Δ) must be positive.
Therefore, Δ > 0:
k^2 - 8k + 4 > 0
To solve this inequality, we need to find the range of possible values for k. Here's how you can do it:
1. Factor the quadratic expression:
(k - 4)^2 > 12
2. Take the square root of both sides:
k - 4 > √12
k - 4 > 2√3
3. Add 4 to both sides:
k > 4 + 2√3
Thus, the possible values for k that yield real and positive roots for the given quadratic equation are k > 4 + 2√3 (approximately 7.464).
Therefore, k must be greater than approximately 7.464 for the quadratic equation to have real and positive roots.
b^2=4ac
so,
(k+2)^2=-12k
k^2+4+4k=-12k
k^2+16k+4=0