Question 6535: What describes the number and type of roots of the equation 4x^3 - 6x^2 + 15x + 2= 0?
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Let f(x) = 4x^3 - 6x^2 + 15x + 2= 0
f'(x) = 12x^2 - 12x + 15 = 3(4x^2 - 4x + 5)
By the quadratic formula, we that f'(x) has no real roots(since
b^2 - 4ac < 0)
This shows that the function has no critical values and hence
no relative extremas.
By Roll's Theorem if f(x) = 0 has three roots, then f'(x) = 0
will have two roots.
But f is a real polynomial of odd degree, f(x) = 0 must have a
real root. In fact f(0) = 2 >0 and f(-1) = -4 -6 -15+2 < 0.
So, there is a root between 0 and 1. While the other
two roots are conjugate complex numbers.
If you are not above calculus level, then it is very hard to
solve this question. {may use Cardan's formula]
Kenny
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