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In the equation x^2+bx+c=0, b and c are constants, and x is

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In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink]

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In the equation \(x^2+bx+c=0\), b and c are constants, and x is a variable. If m and k are the roots of \(x^2+bx+c=0\), then m−k=?

(1) c=−10
(2) b=3
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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink]

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idinuv wrote:
In the equation \(x^2+bx+c=0\), b and c are constants, and x is a variable. If m and k are the roots of \(x^2+bx+c=0\), then m−k=?

(1) c=−10
(2) b=3


Even when we combine the statements we get \(x^2+3x-10=0\) --> \((x+5)(x-2)=0\) --> \(x=-5\) or \(x=2\) --> m and k are -5 and 2, but we don't know which one is which --> m-k is either -5-2=-7 or 2-(-5) = 7.

Answer: E.
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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink]

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New post 31 Dec 2015, 19:32
tough one..solved it by plugging numbers:
1. suppose k=-2 and m=5. km=-10. now m-k = 5+2=7.
suppose k=5 and m=-2 -> km=-10. now m-k = -2-5=-7.

A is insufficient, and we can cross out A and D.

2. b=3.
m+k=3. now, this alone doesn't tell us much.
m=2 and k=1 and m-k=1 or m=1 and k=2, and thus m-k=-1.
statement 2 alone is not sufficient.

we are left with C and E.

1+2
mk=-10
m+k=3
m=-2 and k=5, mk=-10 and m+k=3 -> m-k = -2-5=-7.
or
m=5 and k=-2, mk=-10, and m+k=3. -> m-k=5+2=7.

2 outcomes, thus both statements are insufficient, and the answer is E.
Re: In the equation x^2+bx+c=0, b and c are constants, and x is   [#permalink] 31 Dec 2015, 19:32
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