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Let \(F = 36(2x - \frac{1}{2})^3\) and \(G = 16(3x - \frac{3}{4})^3\). Which of the following is true for all real numbers x?
(A) \(2F - 3G = 0\)
(B) \(3F - 2G = 0\)
(C) \(F^2 - G^2 = 0\)
(D) \(2F^2 - 3G^2 = 0\)
(E) \(3F^2 - 2G^2 = 0\)
(A) \(2F - 3G = 0\)
(B) \(3F - 2G = 0\)
(C) \(F^2 - G^2 = 0\)
(D) \(2F^2 - 3G^2 = 0\)
(E) \(3F^2 - 2G^2 = 0\)
16 Oct 2023, 15:05
Kudos
Bookmarks
Since x can be any real number, choosing a convenient value for x, let x be 0.
This simplifies F & G as follows:
\(F = 36(-\frac{1}{8}) = -4.5\)
\(G = 16(-\frac{27}{64}) = -6.75\)
Upon cycling through answer choices, \(3F - 2G = 0\) satisfies the solution.
(B)
I was wondering if this is the most optimal solution, or there are other ways to evaluate these types of inequality/algebra questions. Choosing a value for x seems a bit arbitary and prone to error as picking x = 1/4 may have made it trickier. I feel like I got lucky by picking x = 0 and it made things work out. Any help / inputs towards a more robust and efficient method of solving these types of problems will be greatly appreciated!
Update - Ok I just figured an Alternative solution:
Notice \(3x - \frac{3}{4} = \frac{3}{2}(2x - \frac{1}{2})\)
Given the answer choices are after a relationship between F & G, evaluating the ratio \(\frac{F}{G}\)
Let \((2x - \frac{1}{2}) = a\)
\(\frac{F}{G} = \frac{36*a^3 }{ 16*(3a/2)^3} = \frac{2}{3}\)
=> \(3F = 2G \)
=> \(3F - 2G = 0 \)
(B)
I think this might be the appropriate way to solve the problem
This simplifies F & G as follows:
\(F = 36(-\frac{1}{8}) = -4.5\)
\(G = 16(-\frac{27}{64}) = -6.75\)
Upon cycling through answer choices, \(3F - 2G = 0\) satisfies the solution.
(B)
I was wondering if this is the most optimal solution, or there are other ways to evaluate these types of inequality/algebra questions. Choosing a value for x seems a bit arbitary and prone to error as picking x = 1/4 may have made it trickier. I feel like I got lucky by picking x = 0 and it made things work out. Any help / inputs towards a more robust and efficient method of solving these types of problems will be greatly appreciated!
Update - Ok I just figured an Alternative solution:
Notice \(3x - \frac{3}{4} = \frac{3}{2}(2x - \frac{1}{2})\)
Given the answer choices are after a relationship between F & G, evaluating the ratio \(\frac{F}{G}\)
Let \((2x - \frac{1}{2}) = a\)
\(\frac{F}{G} = \frac{36*a^3 }{ 16*(3a/2)^3} = \frac{2}{3}\)
=> \(3F = 2G \)
=> \(3F - 2G = 0 \)
(B)
I think this might be the appropriate way to solve the problem
27 Oct 2023, 05:21
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MrWhite
Let F = \(36(2x - \frac{1}{2})^3\) and G = \(16(3x - \frac{3}{4})^3\). Which of the following is true for all real numbers x?
(A) \(2F - 3G = 0\)
(B) \(3F - 2G = 0\)
(C) \(F^2 - G^2 = 0\)
(D) \(2F^2 - 3G^2 = 0\)
(E) \(3F^2 - 2G^2 = 0\)
(A) \(2F - 3G = 0\)
(B) \(3F - 2G = 0\)
(C) \(F^2 - G^2 = 0\)
(D) \(2F^2 - 3G^2 = 0\)
(E) \(3F^2 - 2G^2 = 0\)
Two ways I could think of immediately
1. Substitute value for x
Let x=0
F = \(36(2x - \frac{1}{2})^3\) = \(36(- \frac{1}{2})^3=\frac{36}{-8}=\frac{-9}{2}\)
G = \(16(3x - \frac{3}{4})^3\) = \(16(- \frac{3}{4})^3=\frac{16*-27}{64}=\frac{-27}{4}=\frac{3*-9}{2*2}=\frac{3F}{2}\)
Or 2G=3F
2. Solve Algebraically
F = \(36(2x - \frac{1}{2})^3\)
Take out 2 from bracket => \(36*2^3(x-\frac{1}{4})^3 \)
Let \((x-\frac{1}{4})^3=a\), so F=36*8a…..(i)
G = \(16(3x - \frac{3}{4})^3\)
Take out 3 from bracket => \(16*3^3(x- \frac{1}{4})^3=16*27a\)……(ii)
Divide i by ii
\(\frac{F}{G}=\frac{36*8a}{16*27a}=\frac{2}{3}\)
3F=2G
B
