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# In the second-degree equation x^2-14x+m = 0, the constant m is a posit

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In the second-degree equation x^2-14x+m = 0, the constant m is a posit  [#permalink]

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18 Feb 2019, 06:54
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75% (hard)

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47% (02:07) correct 53% (02:09) wrong based on 96 sessions

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GMATH practice exercise (Quant Class 16)

In the second-degree equation x^2-14x+m = 0, the constant m is a positive integer. If A<B are the roots of this equation, what is the value of B-A ?

(1) m=33
(2) A and B are two primes

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In the second-degree equation x^2-14x+m = 0, the constant m is a posit  [#permalink]

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18 Feb 2019, 11:24
Given that A & B are the two roots of quadratic equation X^2-14X+m=0 and B>A.

using 1st statement, we get the equation X^2-14X+33=0, which gives us two roots, B=11 and A=3. So B-A=8, definite answer.
using 2nd statement, we can say that both roots, A&B, are prime numbers. We also know that in a quadratic equation, sum of roots=-b/a, which gives us A+B=14. Hence B=11 and A=3. Again we have definite answer.

Since each of the statements individually gives us definite value of B-A, answer is D.
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Re: In the second-degree equation x^2-14x+m = 0, the constant m is a posit  [#permalink]

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18 Feb 2019, 18:23
1
fskilnik wrote:
GMATH practice exercise (Quant Class 16)

In the second-degree equation x^2-14x+m = 0, the constant m is a positive integer. If A<B are the roots of this equation, what is the value of B-A ?

(1) m=33
(2) A and B are two primes

$${x^2} - 14x + m = 0\,\,,\,\,\,m \ge 1\,\,{\mathop{\rm int}}$$

$$? = B - A\,\,\,\,\left( {A < B\,\,{\rm{roots}}} \right)\,\,\left( * \right)$$

$$\left( 1 \right)\,\,{\rm{equation}}\,\,{\rm{known}}\,\,\,\,\, \Rightarrow \,\,\,{\rm{the}}\,\,2\,\,{\rm{roots}}\,\,A,B\,\,\,{\rm{known}}\,\,\,\,\mathop \Rightarrow \limits^{A\, < \,B} \,\,\,\,\,A\,\,{\rm{unique}}\,,B\,\,{\rm{unique}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.$$

$$\left( 2 \right)\,\,\left\{ \matrix{ A + B = 14 \hfill \cr A < B\,\,{\rm{primes}} \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{inspection}}} \,\,\,\,\,\left( {A,B} \right) = \left( {3,11} \right)\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.$$

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: In the second-degree equation x^2-14x+m = 0, the constant m is a posit  [#permalink]

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18 Feb 2019, 19:55
1. condition is already sufficient .

2. second one we have , a+b = 14

which gives two set of values, 7,7 and 11,3

Now ,it is given that a<b , so 7,7 is ruled out .

Hence the answer is D .

Each one iw sufficient

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Re: In the second-degree equation x^2-14x+m = 0, the constant m is a posit  [#permalink]

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06 Jun 2019, 10:25
The previously posted solutions regarding statement 2 seem to overlook the fact that 1,13 works as well. This is confirmed by plugging each into the equation. Based on this logic, statement 2 comes up with 2 solutions: 13-1=12 OR 11-3=9. Therefore, the answer should be 'A'. Please review and let me know if/where I made a mistake.
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Re: In the second-degree equation x^2-14x+m = 0, the constant m is a posit  [#permalink]

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07 Jun 2019, 05:23
mafasano421 wrote:
The previously posted solutions regarding statement 2 seem to overlook the fact that 1,13 works as well. This is confirmed by plugging each into the equation. Based on this logic, statement 2 comes up with 2 solutions: 13-1=12 OR 11-3=9. Therefore, the answer should be 'A'. Please review and let me know if/where I made a mistake.

Hi, mafasano421.

Thank you for your interest in our question/solution.

The number 1 is not prime.

Regards,
Fabio.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Re: In the second-degree equation x^2-14x+m = 0, the constant m is a posit   [#permalink] 07 Jun 2019, 05:23
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