+0  
 
0
30
2
avatar

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.

 Apr 20, 2021
 #1
avatar+496 
+1

The only way it can equal -n is if x itself is negative. we can expand x(x+5) into x^2+5x=-n. if -n has to be an integer, then so does x.we can count all the possible ways:

x=-1
x^2+5x=-4

x=-2
x^2+5x=-6

x=-3
x^2+5x=-6
x=-4
x^2+5x=-4

x=-5

x^2+5x=0

from this point on, there will be no possible solutions.

so the only numbers that has real solutions that any x(x+5)=-n is n=4 or n=6, which is 2/10, so the answer is 8/10=$\boxed{\dfrac45}$

 Apr 20, 2021
 #2
avatar+118505 
+1

x ( x + 5) =  -n

 

x(x + 5)  +  n  =  0

 

x^2  +  5x  +  n  =  0

 

There will be  no real solutions  to this if

 

5^2  -  4n  <  0

 

25  <  4n

 

n > 25/4

 

n >  6.25

 

So....the  integer  n's  that  make  this  true  are 7, 8 , 9  and 10

 

The probability of  a non-real solution   is   4 / 10  =  2  / 5

 

 

cool cool cool

 Apr 20, 2021

16 Online Users

avatar