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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 + 4)(x + 5) = n? Express your answer as a common fraction.

 May 31, 2021
 #1
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(x + 4)(x + 5) =   n        simplify  as

 

x^2  +  9x  + 20 - n   =   0

 

This will  have   real solutions  if  the disxriminant  is  ≥  0.....so.......

 

9^2  -  4 (20 - n) ≥    0

 

81  - 80  +  4n  ≥  0

 

1  +  4n   ≥     0

 

4n  ≥  -1

 

n ≥ -1/4

 

In other words.......any   positive   n   will give real solutions....so....the probability  of non-real solutions =  0%

 

 

cool cool cool

 Jun 1, 2021

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