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.

~ Question from AoPS ~

Background Information
Subject: Algebra
Focus: Percents
*Review Problem

1 Answer
Somebody N.
Jun 14, 2018

#color(blue)(2/5)#

Explanation:

#x(x+5)=-n#

Expand:

#x^2+5x+n=0#

We need the solutions to this equation to be non-real.

Using the discriminant of a quadratic:

#bb(b^2-4ac)#

If:

#b^2-4ac> 0# The equation has two real solutions.

#b^2-4ac= 0# The equation has repeated real solutions.

#b^2-4ac< 0# The equation has non-real solutions.

We need the last situation.

From equation:

#(5)^2-4(1)(n)<0#

#25-4n<0#

#n>25/4#

So for the equation to have no real solutions:

#n>24/4=6.25#

We are choosing integers so n would have to be:

#7, 8 , 9, 10#

So for the probability of this we have:

#("favourable outcomes")/("all possible outcomes")=4/10=2/5#

Related questions
Impact of this question
3033 views around the world
You can reuse this answer
Creative Commons License