+0  
 
+1
51
6
avatar+604 

I choose a random integer  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.

mathtoo  Aug 11, 2018
 #1
avatar+88871 
+2

x ( x + 5)  = -n     add n to  both sides

 

x^2 + 5x + n   = 0

 

This will have no real solutions when the disciminant of the quadratic  is  < 0

 

So we have

 

5^2  - 4 (1)(n) < 0

25 - 4n < 0

25 < 4n

25 / 4 < n  ⇒    n > 6.5

So  n = 7, 8, 9, 10   mean that we have no real solutions for those n's

 

Then....the probability that there are  no real solutions  is  4/10  

 

cool cool cool

CPhill  Aug 11, 2018
 #3
avatar+139 
+1

Great solution, CPhill! 

azsun  Aug 11, 2018
 #4
avatar+88871 
+1

Thanks   !!!!

 

cool cool cool

CPhill  Aug 11, 2018
 #2
avatar+139 
+1

 First we find the solution set that results in the equation having no real solutions. We begin by rearranging the equation \(x(x+5) = -n\) to \(x^2 + 5x + n = 0\). If the discriminant \(x^2 + 5x + n = 0\), then there are no real solutions. Thus, we want to solve for \(n\) in the inequality \(25 - 4n < 0\). Adding  \(4n\)and dividing by 4, we find \(n>6.25\). The probability that I chose one of the numbers 7, 8, 9, or 10 is \(\boxed{\frac{2}{5}}\) .

azsun  Aug 11, 2018
 #5
avatar+604 
+2

Thank you, both CPhill and azsun for wonderful solutions!

mathtoo  Aug 11, 2018
 #6
avatar+88871 
+1

OK, mathtoo    !!!!!

 

 

cool cool cool

CPhill  Aug 11, 2018

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