Determine the range of the values of k at which the quadratic equation 3x^2+8x+2k=0 will have 2 different negative real number solutions.
To find the range of values of k at which the quadratic equation 3x^2 + 8x + 2k = 0 will have 2 different negative real number solutions, we need to use the discriminant.
The discriminant (denoted as Δ) of a quadratic equation ax^2 + bx + c = 0 is given by the formula: Δ = b^2 - 4ac.
For our equation 3x^2 + 8x + 2k = 0, the coefficients are: a = 3, b = 8, and c = 2k.
We need to find the range of values of k for which the discriminant is positive (Δ > 0), which ensures the equation has 2 different real solutions.
Substituting the coefficients into the discriminant formula, we get: Δ = (8)^2 - 4(3)(2k) = 64 - 24k.
For 2 different negative real number solutions, we must also ensure that the roots are real and negative. This happens when the quadratic equation has a positive leading coefficient (a > 0) and the x-coordinate of the vertex is negative.
In our case, the leading coefficient is positive (a = 3). The x-coordinate of the vertex is given by the formula: x = -b / (2a), which, in this case, is -8 / (2 * 3) = -4/3.
Thus, we need to ensure that the vertex (-4/3, y) lies in the negative x-axis.
To do this, we check whether -4/3 is smaller than the x-intercept of the quadratic.
The x-intercepts are found by equating the equation to zero: 3x^2 + 8x + 2k = 0. By factoring or using the quadratic formula, we can find the x-intercepts.
Since we are looking for two different negative real number solutions, we can set the discriminant greater than 0 (Δ > 0):
64 - 24k > 0
Simplifying the inequality, we have:
24k < 64
Dividing both sides of the inequality by 24:
k < 64/24
Simplifying further, we get:
k < 8/3
Therefore, the range of values for k, at which the quadratic equation 3x^2 + 8x + 2k = 0 will have 2 different negative real number solutions, is k < 8/3.
To find the range of values of k for which the quadratic equation 3x^2 + 8x + 2k = 0 has two different negative real number solutions, we need to determine the discriminant of the equation.
The discriminant (Δ) of a quadratic equation of the form ax^2 + bx + c = 0 is given by the formula Δ = b^2 - 4ac. For the equation 3x^2 + 8x + 2k = 0, the discriminant is Δ = (8^2) - 4(3)(2k).
To have two different negative real number solutions, the discriminant must be greater than zero, as this indicates two real roots. Additionally, the roots must be negative, meaning their values are less than zero.
Let's set up the inequalities to solve for k:
Δ > 0 and the roots are negative, so:
(8^2) - 4(3)(2k) > 0 (inequality 1)
and
-8 / (2*3) > 0 (inequality 2)
Simplifying inequality 1:
64 - 24k > 0
Divide both sides by 8:
8 - 3k > 0
Rearranging:
3k < 8
k < 8/3
Simplifying inequality 2:
-4/3 > 0
For both inequalities to be true, we need to consider the maximum value of k that satisfies both conditions, which is k < 8/3.
Therefore, the range of values for k is k < 8/3 to have the quadratic equation 3x^2 + 8x + 2k = 0 with two different negative real number solutions.
the discriminant is 64-24 = 40
since it is positive, there will be two real roots for all values of k.
Since we want negative roots, k must be positive.
since the roots are different, consider
3(x^2 + 8/3 x + 16/9) + 2k - 3*16/9 = 0
3(x^2 + 4/3)^2 + 2k-16/3 = 0
so 2k > 16/3
k > 8/3
check:
k=2: 3x^2+8x+4 = (3x+2)(x+2) has two different negative roots
k=8/3: a repeated root at x = -4/3
k=3: 3x^2+8x+6 has no real roots