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Algebra

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13 Apr 2007 21:32:00 IST

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590

Brain cracker!!!!!!!!

None

Each coefficient of the equation ax²+bx+c=0 is detemined by throwing an ordinary dice.The probability that the equation has non real complex roots is

(a) 173/216 (b)43/216 (c)54/216 (d)none of these

Ans is (a)173/216

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Amar Gupta

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Joined: 31 Jan 2007

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13 Apr 2007 22:01:20 IST

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Dear,

1<=a,b,c<=6

for complex roots : b^2-4ac<0

1) now for b=1, 1-4ac<0

for every possible value of a and c eq. is true so favourable cases=6*6 =36

2) for b=2, 4-4ac<0

a=1 and c=1 this will not satisfy. so (1,1) will not satisfy.

so favourable cases=36-1=35

3) for b=3, 9-4ac<0

(1,1) (1,2) (2,1) will not satisfy.

so favourable cases=36-3=33

4) for b=4, 16-4ac<0

(1,1) (1,2) (1,3) (1,4) (2,1) (2,2) (3,1) (4,1) will not satisfy.

so favourable cases=36-8=28

5) for b=5, 25-4ac<0

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (3,1) (3,2) (4,1) (5,1) (6,1) will not satisfy.

so favourable cases=36-14=22

6) for b=6, 36-4ac<0

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (3,1) (3,2) (3,3) (4,1) (4,2) (5,1) (6,1) will not satisfy.

so favourable cases=36-17=19

total faurable cases (P) = 36+35+33+28+22+19 = 173

total ways possible (Q) = 6*6*6 =216

probability = P/Q = 173/216

Siddharth Agarwal

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Joined: 9 Apr 2007

Posts: 166

13 Apr 2007 22:46:45 IST

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This question is just a mechanical one. Not really a "brain cracker".

CyBorG

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Joined: 6 Jan 2007

Posts: 706

14 Apr 2007 09:05:09 IST

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Thank you very much sir!
I made a small mistake while calculating the no. of favourable cases.
And sidlol dont take the headings given to the problem so seriously!!!!!!!!!!!!!!

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