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Each coefficient in the equation ax2 + bx + c = 0 is determined by thr
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19 Jun 2019, 17:45
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42% (01:57) correct 58% (02:36) wrong based on 38 sessions
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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
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Each coefficient in the equation ax2 + bx + c = 0 is determined by thr
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Updated on: 20 Jun 2019, 02:39
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)
Originally posted by firas92 on 19 Jun 2019, 18:52.
Last edited by firas92 on 20 Jun 2019, 02:39, edited 1 time in total.



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Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr
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20 Jun 2019, 02:23
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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Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr
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20 Jun 2019, 02:35
omkartadsare wrote: Hi, can you please explain why the coefficients can't be B=6,a=3 and C=3, for ax^2+bx+c
Posted from my mobile device You are right! Thanks for pointing that out.. I fell into a classic trap! {2,4,2} and {3,6,3} also work. I suppose the answer should be 5/216 then!



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Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr
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20 Jun 2019, 07:31
Dividing the equation by a, we get x^2 + (b/a)x + (c/a) = 0, which must also satisfy (x+√ (c/a))^2 = 0 in order for the equation to have equal roots.
Expanding the 2nd equation gets x^2 + 2√ (c/a)x + (c/a) = 0, and comparing its 2nd term to the first equation's 2nd term, we get b/a = 2√ (c/a). Since a, b and c must be integers from 1 to 6, we can square and manipulate the equation to get (b/a)^2 = 4(c/a) b^2 = 4ac b^2  4ac = 0
b^2  4ac is the discriminant in the quadratic formula, and although GMAT doesn't require the knowledge of the quadratic formula, it certainly makes sense for it to equal 0 here in order to get equal roots.
Now, to calculate the probability of this equation holding, test with real values to see how many work. b=2 > a=1, c=1 b=4 > a=1, c=4 b=4 > a=4, c=1 b=4 > a=2, c=2 b=6 > a=3, c=3
So there are 5 combinations that work out of a total combination of 6*6*6 = 216 rolls.
Therefore the probability needed is 5/216. The answer is C.



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Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr
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09 Aug 2019, 12:53
Is there a way to solve this without having to test real values? I feel it would take me 62 minutes to just test them all... I didn't connect 'equal roots' with b^24ac, great explanation



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Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr
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09 Aug 2019, 18:20
It shouldn't take long to test with real values because the values can be derived.
4ac is even so b^2 has to be even, hence b has to be even. b=2: 4=4ac 1=ac
b=4: 16=4ac 4=ac
b=6: 36=4ac 9=ac
It's just a matter of solving these 3 equations. Hope this helps.
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Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr
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09 Aug 2019, 18:20






