determine the value(s) of k which the equation x^2+kx+9=0 will have
a) two real roots
b) 2 distinct roots
I tried to find the discriminant but I didn't get the right answer.
why not show your work?
The discriminant is k^2-36
Now, what does that tell you about the number of roots?
To determine the value(s) of k for which the equation x^2 + kx + 9 = 0 has two real roots or two distinct roots, we need to consider the discriminant of the quadratic equation.
The discriminant, denoted as Δ, is given by the formula Δ = b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.
In this case, the quadratic equation is x^2 + kx + 9 = 0, so we have a = 1, b = k, and c = 9.
To find the value(s) of k for which the equation has two real roots:
For a quadratic equation to have two real roots, the discriminant must be greater than 0 (Δ > 0).
Thus, we have:
Δ = b^2 - 4ac > 0
k^2 - 4(1)(9) > 0
k^2 - 36 > 0
Now, we solve this inequality to find the range of values for k:
k^2 > 36
k > √36
k > ±6
Therefore, the equation x^2 + kx + 9 = 0 will have two real roots when k > 6 or k < -6.
To find the value(s) of k for which the equation has two distinct roots:
For a quadratic equation to have two distinct roots, the discriminant must be greater than 0 (Δ > 0), and the equation must not have any repeated roots.
Since we already established that for two real roots, k > 6 or k < -6, we need to check that there are no repeated roots within this range.
To ensure no repeated roots, we also need to consider the case when the discriminant is equal to 0 (Δ = 0).
Thus, we have:
Δ = b^2 - 4ac = 0
k^2 - 36 = 0
k^2 = 36
k = ±√36
k = ±6
Therefore, the equation x^2 + kx + 9 = 0 will have two distinct roots when k > 6 or k < -6, excluding the value k = 6 or k = -6.
In summary:
a) The equation x^2 + kx + 9 = 0 will have two real roots when k > 6 or k < -6.
b) The equation x^2 + kx + 9 = 0 will have two distinct roots when k > 6 or k < -6, excluding the value k = 6 or k = -6.