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# Help.....

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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.

Aug 11, 2018

#1
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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   Aug 11, 2018
#3
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Great solution, CPhill!

azsun  Aug 11, 2018
#4
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Thanks   !!!!   CPhill  Aug 11, 2018
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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 $$\boxed{\frac{2}{5}}$$ .

Aug 11, 2018
#5
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Thank you, both CPhill and azsun for wonderful solutions!

Aug 11, 2018
#6
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OK, mathtoo    !!!!!   CPhill  Aug 11, 2018