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# halp

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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.

Apr 20, 2021

#1
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The only way it can equal -n is if x itself is negative. we can expand x(x+5) into x^2+5x=-n. if -n has to be an integer, then so does x.we can count all the possible ways:

x=-1
x^2+5x=-4

x=-2
x^2+5x=-6

x=-3
x^2+5x=-6
x=-4
x^2+5x=-4

x=-5

x^2+5x=0

from this point on, there will be no possible solutions.

so the only numbers that has real solutions that any x(x+5)=-n is n=4 or n=6, which is 2/10, so the answer is 8/10=\$\boxed{\dfrac45}\$

Apr 20, 2021
#2
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x ( x + 5) =  -n

x(x + 5)  +  n  =  0

x^2  +  5x  +  n  =  0

There will be  no real solutions  to this if

5^2  -  4n  <  0

25  <  4n

n > 25/4

n >  6.25

So....the  integer  n's  that  make  this  true  are 7, 8 , 9  and 10

The probability of  a non-real solution   is   4 / 10  =  2  / 5   Apr 20, 2021