3x2 - 3kx + k2 > 0, k = ? (a = 3, b = -3k, c = k2)
The parabola opens upward (a > 0), so we have a minimum point at the
vertex = (- b/2a, c - b2/4a) = (- -3k/6, k2 - (-3k)2/12) = (k/2, k2/4).
Since the y-coordinate of the vertex is always a positive value, except when k = 0, then the x-coordinate could have a positive or negative value.
So that, the parabola would lie above the x-axis for all values of k, except when k is zero.
Thus, the equation 3x2 - 3kx + k2 > 0, for k ≠ 0.
Or
Complete the square:
3x2 - 3kx + k2 (divide by 3 all the terms)
= x2 - kx + k2/3
= [x2 - kx + (k/2)2]+ k2/3 - (k/2)2
= (x - k/2)2 + k2/3 - k2/4
= (x - k/2)2 + k2/12
So the equation (x - k/2)2 + k2/12 represents the translation of x2, k/2 units right when k > 0, or k/2 units left when k < 0, and k2/12 units up.
Thus, for k ≠ 0 the given equation is always positive.
No. Probability values always have to be positive.
An absolute value doesn't have 2 answers. It isn't even a question. It's always positive.
an equation that's true for all values is an identity.
For the equation: x2+4x-5 = 0 Factoring yields: (x+5)(x-1) = 0 The zero identity states that setting these factors individually equal to zero will yield valid solutions to this equation. x+5 = 0, x = -5 x-1 = 0, x = 1 So the values x=-5 and x=1 solve this equation. For the mathematical function: y = x2+4x-5 The derivatives for this function are as such: y' = 2x+4 y'' = 2 The second derivative is a constant positive number (2), so it is always positive. Since the second derivative is always positive, its curvature is always concave upward.
Definitely.The equation [ x^2 = 4 ] has two solutions.x = +2x = -2The square root of any number can be a positive number or its negative. The solution for a quadratic equation often has two different values. However having two different values is still a single solution.
No. Probability values always have to be positive.
An equation that is always true is an identity.
No, absolute values are always positive.
yes.
For positive numbers 8,6,4,2
It is true for all permissible values of any variables in the equation. More simply put, it is always true.
Absolute values are always positive; so graph it on the positive side of the number line.
An absolute value doesn't have 2 answers. It isn't even a question. It's always positive.
The domain of a parabola is always all real numbers because the domain represents the possible x values. The x values are shown on the horizontal axis or x axis. Because, in a parabola, the 2 sides of the parabola go infinitely in a positive or negative direction, there is always a y value for any x value that u plug in to the equation.
an equation that's true for all values is an identity.
Find values for the variable that satisfy the equation, that is if you replace those values for the variable into the original equation, the equation becomes a true statement.
For the equation: x2+4x-5 = 0 Factoring yields: (x+5)(x-1) = 0 The zero identity states that setting these factors individually equal to zero will yield valid solutions to this equation. x+5 = 0, x = -5 x-1 = 0, x = 1 So the values x=-5 and x=1 solve this equation. For the mathematical function: y = x2+4x-5 The derivatives for this function are as such: y' = 2x+4 y'' = 2 The second derivative is a constant positive number (2), so it is always positive. Since the second derivative is always positive, its curvature is always concave upward.