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f(x)=Ax^2+Bx+7
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22 Mar 2024, 04:48
22 Mar 2024, 04:48
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Question Stats:
83% (02:56) correct
16% (02:09) wrong based on 12 sessions
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\(f(x)=Ax^2+Bx+7\)
\(f(3)=34\)
\(f(4)=51\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
\(f(3)=34\)
\(f(4)=51\)
Quantity A |
Quantity B |
A |
B |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Part of the project: The Butler-GRE Daily New Quant and Verbal Questions to Practice (2023) - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
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f(x)=Ax^2+Bx+7
[#permalink]
10 May 2024, 12:33
10 May 2024, 12:33
1
Find f(3) and f(4) in terms of A and B:
\(f(3) = A(3)^2 + 3B + 7 = 34\)
\(9A + 3B = 27\)
\(f(4) = A(4)^2 + 4B + 7 = 51\)
\(16A + 4B = 44\)
Use systems of equations to find A:
\(3(16A+4B)-4(9A + 3B) = 3(44)-4(27)\)
\(48A + 12B - 36A - 12B = 24\)
\(12A = 24\)
\(A = 2\)
Plug A = 2 back into 9A + 3B = 27 to find B:
\(9(2) + 3B = 27\)
\(3B = 9\)
\(B = 3\)
Since 3>2, B is greater than A.
\(f(3) = A(3)^2 + 3B + 7 = 34\)
\(9A + 3B = 27\)
\(f(4) = A(4)^2 + 4B + 7 = 51\)
\(16A + 4B = 44\)
Use systems of equations to find A:
\(3(16A+4B)-4(9A + 3B) = 3(44)-4(27)\)
\(48A + 12B - 36A - 12B = 24\)
\(12A = 24\)
\(A = 2\)
Plug A = 2 back into 9A + 3B = 27 to find B:
\(9(2) + 3B = 27\)
\(3B = 9\)
\(B = 3\)
Since 3>2, B is greater than A.
gmatclubot
f(x)=Ax^2+Bx+7 [#permalink]
10 May 2024, 12:33
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