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Each coefficient in the equation ax2 + bx + c = 0 is determined by thr

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Each coefficient in the equation ax2 + bx + c = 0 is determined by thr  [#permalink]

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New post 19 Jun 2019, 17:45
2
8
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

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  [#permalink]

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New post Updated on: 20 Jun 2019, 02:39
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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)

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  [#permalink]

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New post 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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Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr  [#permalink]

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New post 20 Jun 2019, 02:35
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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

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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  [#permalink]

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New post 20 Jun 2019, 07:31
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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  [#permalink]

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New post 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^2-4ac, great explanation
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Re: Each coefficient in the equation ax2 + bx + c = 0 is determined by thr  [#permalink]

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New post 09 Aug 2019, 18:20
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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   [#permalink] 09 Aug 2019, 18:20
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