For what real values of k will the roots of the equation kx^2-3kx=2 be real numbers?
k x^2 -3k x - 2 = 0
a = k
b = -3k
c = -2
look at b^2 - 4 a c
look at parabola y = 9k^2 + 8 k
must be 0 or +
for 0
k (9k+8) = 0
k = 0 or k= -8/9
between them, y is negative
SO
k</= -8/9
or
k>/=0
To find the real values of k for which the roots of the equation are real numbers, we can use the discriminant.
The discriminant, denoted as Δ, is a mathematical expression used to determine the nature of the roots of a quadratic equation. For a quadratic equation in the form ax^2 + bx + c = 0, the discriminant is given by Δ = b^2 - 4ac.
In this case, we have the equation kx^2 - 3kx = 2. Comparing it with the general form of a quadratic equation, we can see that a = k, b = -3k, and c = -2.
Now, let's find the discriminant:
Δ = (-3k)^2 - 4(k)(-2)
= 9k^2 + 8k
For the roots to be real numbers, the discriminant Δ must be greater than or equal to zero. Therefore, we need to solve the inequality 9k^2 + 8k ≥ 0.
To solve this inequality, we can factor out k:
k(9k + 8) ≥ 0
Now we have two cases to consider:
1. k > 0:
In this case, both k and (9k + 8) are positive. So the inequality k(9k + 8) ≥ 0 is satisfied.
2. k < 0:
In this case, k is negative while (9k + 8) is positive. Since the inequality k(9k + 8) ≥ 0 must be satisfied, we have no valid solution for this case.
Thus, the values of k for which the roots of the equation kx^2 - 3kx = 2 are real numbers are k > 0.