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19 Jun 2019, 17:45
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C
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Difficulty:
95%
(hard)
Question Stats:
42% (02:22) correct
58%
(02:33)
wrong
based on 101
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Each coefficient in the equation \(ax^2 + bx + c = 0\) is determined by throwing an ordinary die. Find the probability that the equation will have equal roots.
A. 1/72
B. 1/54
C. 5/216
D. 1/36
E. 1/27
A. 1/72
B. 1/54
C. 5/216
D. 1/36
E. 1/27
firas92
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Joined: 16 Jan 2019
Last visit: 02 Dec 2024
Posts: 616
Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
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Each of the coefficients must be an integer greater than 0 and less than 7
When the equation can be in the form \((mx+n)^2\) we get equal roots
We have
\((x+1)^2=x^2+2x+1\) -----{1,2,1}
\((x+2)^2=x^2+4x+4\) -----{1,4,4}
\((2x+1)^2=4x^2+4x+1\) -----{4,4,1}
In addition we have cases where b=a+c {2,4,2} and {3,6,3}
Total number of possibilities for the coefficients is 6*6*6
Therefore the required probability = \(\frac{5}{6*6*6} = \frac{5}{216}\)
Answer is (C)
When the equation can be in the form \((mx+n)^2\) we get equal roots
We have
\((x+1)^2=x^2+2x+1\) -----{1,2,1}
\((x+2)^2=x^2+4x+4\) -----{1,4,4}
\((2x+1)^2=4x^2+4x+1\) -----{4,4,1}
In addition we have cases where b=a+c {2,4,2} and {3,6,3}
Total number of possibilities for the coefficients is 6*6*6
Therefore the required probability = \(\frac{5}{6*6*6} = \frac{5}{216}\)
Answer is (C)
20 Jun 2019, 02:23
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Hi, can you please explain why the coefficients can't be B=6,a=3 and C=3, for ax^2+bx+c
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