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If x is an integer such that x^5(x + 2)^3(x - 3)^5/(x + 4) <= 0, what

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If x is an integer such that x^5(x + 2)^3(x - 3)^5/(x + 4) <= 0, what  [#permalink]

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New post 02 Jun 2020, 07:19
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If x is an integer such that x^5(x + 2)^3(x - 3)^5/(x + 4) <= 0, what  [#permalink]

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New post 02 Jun 2020, 07:38
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Bunuel wrote:
If x is an integer such that \(\frac{x^5(x + 2)^3(x - 3)^5}{(x + 4)} \leq 0\), what is the probability that \(x^2 -2x - 8 = 0\)?


A. 1/7

B. 1/6

C. 1/4

D. 2/7

E. 1/3


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Asked: If x is an integer such that \(\frac{x^5(x + 2)^3(x - 3)^5}{(x + 4)} \leq 0\), what is the probability that \(x^2 -2x - 8 = 0\)?

Using Wavy line method: -
\(\frac{x^5(x + 2)^3(x - 3)^5}{(x + 4)} \leq 0\)
-4<x<=-2 ; 0<=x=<3
x = {-3,-2,0,1,2,3}

\(x^2 -2x - 8 = 0\)
(x-4)(x+2) = 0
x = {-2}; Since x = 4 is not in the domain of x

Probability = 1/6

IMO B
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If x is an integer such that x^5(x + 2)^3(x - 3)^5/(x + 4) <= 0, what  [#permalink]

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New post 02 Jun 2020, 12:58
IMO B

Integer Values of x satisfying the equation: {-3, -2, 0, 1, 2, 3}

for the eqn x^2 -2x -8 = 8,roots are: -2, 4

Only -2 is in the set

Probability = 1/6
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Re: If x is an integer such that x^5(x + 2)^3(x - 3)^5/(x + 4) <= 0, what  [#permalink]

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New post 06 Jun 2020, 04:23
1
Bunuel wrote:
If x is an integer such that \(\frac{x^5(x + 2)^3(x - 3)^5}{(x + 4)} \leq 0\), what is the probability that \(x^2 -2x - 8 = 0\)?


A. 1/7

B. 1/6

C. 1/4

D. 2/7

E. 1/3



Solution:

We see that if x = 0, -2, or 3, the expression on the left hand side of the inequality will be 0 and thus satisfying the inequality. If x = -4, the denominator of the expression is 0. Since we can’t divide by 0, x can’t be -4. If x < -4, all the factors (x^5, (x + 2)^3, (x - 3)^5 and (x + 4)) are negative. However, the expression will be positive, so it won’t satisfy the inequality. Similarly, if x > 3, all the factors are positive and the expression will be positive.

Now, let’s determine what happens if x is an integer between -4 and 3 (not including 0 and -2).

If x = -3, x^5, (x + 2)^3 and (x - 3)^5 are negative, but (x + 4) is positive. So the expression is negative. Thus, x = -3 is a solution for the inequality.

If x = -1, x^5 and (x - 3)^5 are negative, but (x + 2)^3 and (x + 4) is positive. So the expression is positive. Thus, x = -1 is NOT a solution for the inequality.

If x = 1, (x - 3)^5 is negative, but x^5, (x + 2)^3 and (x + 4) is positive. So the expression is negative. Thus, x = 1 is a solution for the inequality.

If x = 2, (x - 3)^5 is negative, but x^5, (x + 2)^3 and (x + 4) is positive. So the expression is negative. Thus, x = 2 is a solution for the inequality.

Therefore, the integer solutions for the inequality are -3, -2, 0, 1, 2 and 3 (i.e., 6 such integers). Now let’s check the integer solutions of the equation by solving it.

(x - 4)(x + 2) = 0

x = 4 or x = -2

Since only -2 is a solution of the inequality, we see that the desired probability is 1/6.

Answer: B


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Re: If x is an integer such that x^5(x + 2)^3(x - 3)^5/(x + 4) <= 0, what   [#permalink] 06 Jun 2020, 04:23

If x is an integer such that x^5(x + 2)^3(x - 3)^5/(x + 4) <= 0, what

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