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.

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Subject: Algebra
Focus: Percents
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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

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