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10 Oct 2008, 10:04
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If b, c, and d are constants and x^2 + b^x + c = (x+d)^2 for all values of x, what is the value of c?
(1) d = 3
(2) b = 6
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
x^2 +b^x+c=x^2+2xd+d^2 bx-2xd=d^2-c x( b-2d)=d^2-c
Stm1 d=3 x(b-6)=9-c insufficient
Stm2 b=6 x(6-2d)=d^2-c insufficient
Combine stm1 and stm2
c=3 the answer is C. Is that correct?
(1) d = 3
(2) b = 6
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
x^2 +b^x+c=x^2+2xd+d^2 bx-2xd=d^2-c x( b-2d)=d^2-c
Stm1 d=3 x(b-6)=9-c insufficient
Stm2 b=6 x(6-2d)=d^2-c insufficient
Combine stm1 and stm2
c=3 the answer is C. Is that correct?
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10 Oct 2008, 11:12
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I think the answer is (E)
you have 4 unknowns
b,c,d,x
even if you know what d,b are then you are left with
\(x^2+6^x+c = (x+3)^2\)
\(x^2+6^x+c = x^2+6x+9\)
\(6^x+c=6x+9\)
cannot be solved
x can be 1 and c can be 9 or x can be 2 and c can be -15
you have 4 unknowns
b,c,d,x
even if you know what d,b are then you are left with
\(x^2+6^x+c = (x+3)^2\)
\(x^2+6^x+c = x^2+6x+9\)
\(6^x+c=6x+9\)
cannot be solved
x can be 1 and c can be 9 or x can be 2 and c can be -15
10 Oct 2008, 13:13
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Jcpenny
First, I'm absolutely certain that the question should read:
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
because otherwise it's mathematically impossible that what is stated in the stem is true, unless b = 1, d = 0 and c = -1. Proceeding on that assumption:
We know that
\(x^2 + bx + c = (x+d)^2\)
for all values of x. In particular, it's true for x=0:
\(0^2 + b*0 + c = (0+d)^2 \\\\
c = d^2\)
So you can find c if you know d, and Statement 1 is sufficient.
Further, knowing that \(c = d^2\), we have
\(x^2 + bx + d^2 = (x+d)^2 \\\\
x^2 + bx + d^2 = x^2 + 2dx + d^2 \\\\
bx = 2xd \\\\
b = 2d \\\)
so if you can find b, you can find d, and therefore you can find c. Statement 2 is sufficient.
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
