In the second-degree equation x^2-14x+m = 0 the constant m

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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

Answer: [spoiler]____(D)__[/spoiler]
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by fskilnik@GMATH » Mon Feb 18, 2019 5:13 pm

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fskilnik@GMATH 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
$${x^2} - 14x + m = 0\,\,,\,\,\,m \ge 1\,\,{\mathop{\rm int}} $$
$$? = B - A\,\,\,\,\left( {A < B\,\,{\rm{roots}}} \right)\,\,\left( * \right)$$
$$\left( 1 \right)\,\,{\rm{equation}}\,\,{\rm{known}}\,\,\,\,\, \Rightarrow \,\,\,{\rm{the}}\,\,2\,\,{\rm{roots}}\,\,A,B\,\,\,{\rm{known}}\,\,\,\,\mathop \Rightarrow \limits^{A\, < \,B} \,\,\,\,\,A\,\,{\rm{unique}}\,,B\,\,{\rm{unique}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.$$
$$\left( 2 \right)\,\,\left\{ \matrix{
A + B = 14 \hfill \cr
A < B\,\,{\rm{primes}} \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{inspection}}} \,\,\,\,\,\left( {A,B} \right) = \left( {3,11} \right)\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.$$

The correct answer is (D).


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Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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