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Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
05 Aug 2011, 06:16
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
50% (01:28) correct
50%
(01:54)
wrong
based on 1263
sessions
History
Date
Time
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Not Attempted Yet
If b, c, and d are constants and \(x^2 + bx + c = (x + d)^2\) for all values of x, what is the value of c?
(1) d = 3
(2) b = 6
(1) d = 3
(2) b = 6
05 Aug 2011, 07:32
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Galiya
If b, c, and d are constants and \(x^2 + bx + c = (x + d)^2\) for all values of x, what is the value
of c?
(1) d = 3
(2) b = 6
of c?
(1) d = 3
(2) b = 6
\(x^2 + bx + c = (x + d)^2\)
\(x^2 + bx + c = x^2 + d^2 + 2xd\) {:Note: (a+b)^2=a^2+b^2+2ab}
\(bx + c = d^2 + 2xd\) {:Note: (a+b)^2=a^2+b^2+2ab}
(1) d=3
\(bx + c = 9 + 6x\)
LHS=RHS for all x;
So, for x=0
\(b*0 + c = 9 + 6*0\)
\(c = 9\)
Sufficient.
(2) b=6
\(bx + c = d^2 + 2xd\)
LHS=RHS for all x;
So, for x=0
\(c=d^2\)----------1
for x=1
\(b + c = d^2 + 2d\)---------2
Using 1 and 2:
\(b = 2d\)
\(6=2d\)
\(d=3\)
Using 1:
\(c=d^2=3^2=9\)
Sufficient.
Ans: "D"
General Discussion
05 Aug 2011, 07:21
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Galiya
If b, c, and d are constants and \(x^2 + bx + c = (x + d)^2\) for all values of x, what is the value
of c?
(1) d = 3
(2) b = 6
of c?
(1) d = 3
(2) b = 6
Show SpoilerComment
I got A instead of OA D
could you explain me please why it so?
could you explain me please why it so?
ok here is what i think is the reasoning ...
expand the eqn .. \(x^2 + bx + c = x^2 + 2dx + d^2\)
cancelling the first term and comparing like terms we get
\(bx = 2dx << i.e. >> \\
\\
b=2d\)
and \(c = d^2\)
now check the options we'd get both A and B to suffice ....
