Given that x^2-6x+3k=0 has 2 real number roots, a and B, with opposite signs (one root is positive and the other root is negative) determine the range of values of k.
x^2 - 6x = 0 has two roots of 0 and 6
looking at the graph of y = x^2 - 6x , if I drop the graph I will get 2 roots, one positive and the other negative.
e.g. x^2 - 6x - 16 = 0 will have roots of -2 and 8
so in x^2 - 6x + 3k = 0
3k < 0
k < 0
check by graphing in a webpage like
www.desmos.com/calculator
To determine the range of values of k, we need to consider the discriminant of the quadratic equation. The discriminant (denoted as D) is given by the formula D = b^2 - 4ac, where the quadratic equation is in the form of ax^2 + bx + c = 0.
In this case, the given quadratic equation is x^2 - 6x + 3k = 0, which means a = 1, b = -6, and c = 3k.
Since the equation has two real number roots with opposite signs, it means that the discriminant D must be positive.
D > 0
(b^2 - 4ac) > 0
((-6)^2 - 4(1)(3k)) > 0
(36 - 12k) > 0
36 > 12k
3 > k
Therefore, the range of values of k that satisfy the condition of having 2 real number roots with opposite signs is k less than 3.