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Find the range of values of k which the equation 3x2-3kx plus k2 is always positive?
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.
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