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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.

 May 4, 2022
 #1
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The values that lead to non-real solutions are 9 and 10, so the probability is 2/10 = 1/5.

 May 4, 2022
 #2
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x(x+5) = -n

 

\(x(x+5) = -n\\ x^2-5x+n=0\\ x=\frac{5\pm\sqrt{25-4n}}{2}\\ \)

 

this will NO real solutions if the number under the square root is negative.

 

\(25-4n<0\\ 25<4n\\ 4n>25\\ n>6.25\)

 

So there will be no real solution if n = 7,8,9 or 10

 

that is  4/10 = 2/5 probability

 May 5, 2022
edited by Melody  May 5, 2022

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