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I choose a random integer \(n\) between 1 and 10 inclusive. What is the probability that for the \(n\) I chose, there exist no real solutions to the equation \(x(x+5) = -n\)? Express your answer as a common fraction.

 Jul 15, 2019
 #1
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x(x+5) = -n if and only if x+5 is -1. then x must be -6. The probability must be zero. :) :) :)

 Jul 15, 2019
 #2
avatar+105195 
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x(x + 5)  = -n

 

x^2 + 5x  = -n

 

x^2 + 5x  + n  = 0

 

This will have no real solutions  when

 

5^2  - 4(1)n < 0

 

25 - 4n < 0

 

25  < 4n

 

25/4 < n     this means that this will have no real solutions when

 

n > 6.25

 

So...it will have no real solutions when  n  =  7, 8, 9  or 10

 

So....the probability of no real solutions =  4 /10   =    2 / 5

 

 

cool cool cool

 Jul 15, 2019

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