If the quadratic equation 2 (KX & # 178; + 3) = x (X-8) has no real root, then the minimum integer of K is

2(kx²+3)=x(x-8)
2kx²+6=x²-8x
(2k-1)x²+8x+6=0
Because the equation has no real roots
So △ = 8 & # 178; - 4 × (2k-1) × 6 ＜ 0
The solution is k > 11 / 6
So the smallest integer k is 2
Answer: 2

Let f (x) = - x + 2, X ∈ [- 5,5]. If a real number x is randomly selected from the interval [- 5,5], what is the probability that the real number x satisfies f (x) less than or equal to 0

When f (x) ≤ 0,
-If x + 2 ≤ 0, then x ≥ 2
∵x∈ [-5,5]
The value range of real number x satisfying that f (x) is less than or equal to 0 is [2,5]
The interval length of interval [2,5] is 5-2 = 3
The length of interval [- 5,5] is 5 - (- 5) = 10
P (the real number x satisfies that f (x) is less than or equal to 0) = 3 / 10
That is to say, the probability of real number x satisfying f (x) less than or equal to 0 is 3 / 10

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first of May, two thousand and eleven

If the quadratic equation 2 (KX & # 178; + 3) = x (X-8) has no real root, then the minimum integer of K is