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

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Bunuel
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In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink] New post   02 May 2021, 06:45
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40% (01:46) correct 60% (01:48) wrong based on 97 sessions
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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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Bunuel
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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink] New post   09 May 2021, 04:30
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Bunuel 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] New post   02 May 2021, 11:07
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Difference of roots is given by formula
\(\frac{\sqrt{D}}{a}\)
D = \(b^2-4ac\)

so we need both B and C , given a - 1
thus both stmts i and ii are needed

C IMO
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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink] New post   12 May 2021, 03:48
globaldesi wrote:
Difference of roots is given by formula
\(\frac{\sqrt{D}}{a}\)
D = \(b^2-4ac\)

so we need both B and C , given a - 1
thus both stmts i and ii are needed

C IMO


the square of the difference = \(b^2-4ac\)
since a=1,
the square of the difference = \(b^2-4c\)
using both statements,
the square of the difference = 49
difference = +-7
Hence, E
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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink] New post   12 May 2021, 05:14
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Solution:

St(1):c=−10
Sum of the roots which is m+ k=-b and product mk =c=-10 (Insufficient as we do not know b)

St(2):-b=3

Insufficient as we do not know c here

On combining,
mk = -10

= >m =-10/k

-10/k + k =-3

=>-10 +k^2 +3k = 0

So we shall have two values of k and hence 2 values for m-k. (Insufficient)

(option e)

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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink] New post   07 Jun 2021, 10:05
if we are in a hurry we are definitely gonna miss the prize and opt out for C just look out the explanation :
A provides mk=-10 clearly insuff
B provides m+k = -3 insuff
when combined we get m=5 ,k=-2 or m=-2, k=5 which gives answers 7 and -7 respectively clearly insufficient
therefore imo E
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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink]
07 Jun 2021, 10:05
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Bunuel
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