Probability in roots

Algebra Level pending

p p and q q are randomly chosen (with replacement) from the set of integers from 1 to 10 (inclusive). What is the probability that the roots of the quadratic equation x 2 + p x + q = 0 x^2 + px + q = 0 are real?


The answer is 0.62.

Bibhor Singh by Bibhor Singh , 18, India

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1 solution

Bibhor Singh
Dec 22, 2017

The required probability = 1 - (probability of the event that the roots of x2+px+q= 0 are non-real).
p and q are integers.
The roots will be non-real, iff p2<4q ( p2 = p square).
The possible values of p and q are-


Value of q Value of p No. of pairs of p,q
1 1 1
2 1,2 2
3 1,2,3 3
4 1,2,3 3
5 1,2,3,4 4
6 1,2,3,4 4
7 1,2,3,4,5 5
8 1,2,3,4,5 5
9 1,2,3,4,5 5
10 1,2,3,4,5,6 6

Thus, no. of possible pairs=38
No. of total possible pairs=10*10=100
Therefore,
The reqired probability= 1-(38/100)= 1-0.38 = 0.62 |

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FYI To start a new line, leave 3 empty space at the end. I've edited your solution for your reference.

Calvin Lin Staff - 3 years, 5 months ago

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Heartly thanks to you for making such effort, actually being a beginner and coming from a non-programming background makes it difficult for me to write in such format. BUT THANK YOU A LOT!!!

Bibhor Singh - 2 years, 8 months ago

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