The root of x ^ 2-5x-36 = 0

The root of x ^ 2-5x-36 = 0

x^2-5x-36=0
(x+4)*(x-9)=0
therefore
X + 4 = 0 or X-9 = 0
So x = - 4 or x = 9
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Discuss the root of the equation: x ^ 3-5x-2 = 0, in (0, positive infinity)

This paper discusses the root of the equation: X & # 179; - 5x-2 = 0, in (0, + ∞)
Let f (x) = x & # 179; - 5x-2; since f '(x) = 3x & # 178; - 5 = 3 (X & # 178; - 5 / 3) = 3 [x + √ (5 / 3)] [x - √ (5 / 3)]
When x ≤ - √ (5 / 3) or X ≥ √ (5 / 3), f '(x) ≥ 0, so f (x) in the interval (- ∞, - √ (5 / 3)] ∪ [√ (5 / 3), + ∞)
When - √ (5 / 3) ≤ x ≤ √ (5 / 3), f '(x) ≤ 0, so it is reduced in the interval [- √ (5 / 3), √ (5 / 3)]
Therefore, in the interval (0, + ∞), f (x) has only one minimum point x = √ (5 / 3), and the minimum value = f [√ (5 / 3)]
=[√(5/3)]³-5[√(5/3)]-2=[√(5/3)][(5/3)-5-2]=-(16/3)√(5/3)