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x2 = y2 + 1 and y ≠ 0.
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06 Apr 2018, 15:36
06 Apr 2018, 15:36
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Expert Reply
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Question Stats:
50% (00:57) correct
50% (00:44) wrong based on 80 sessions
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\(x^2 = y^2 + 1\) and y ≠ 0.
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.
Quantity A |
Quantity B |
\(x^4\) |
\(y^4+1\) |
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.
Drill 1
Question: 12
Page: 509
Question: 12
Page: 509
Re: x2 = y2 + 1 and y ≠ 0.
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07 Apr 2018, 07:27
07 Apr 2018, 07:27
You can solve this by addding variables but I think it’s easier, tell me if I’m wrong, to look at y^4 + 1. By definition this will be always greater than x^4. Solution is A.
Re: x2 = y2 + 1 and y ≠ 0.
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18 Apr 2018, 15:39
18 Apr 2018, 15:39
1
Expert Reply
Explanation
Plugging in 2 for y gives you \(x^2 = 5\) in the given equation and 17 for Quantity B.
Squaring this gives you \(x^4 = 25\) for Quantity A, which is therefore larger.
Plugging in any other number gives the same result.
Alternatively, doing algebra by squaring both sides of the given equation reveals Quantity A: \(x4 = (y^2 + 1)(y^2 + 1) = y^4 + 2y^2 + 1\).
The only difference between Quantities A and B is the \(2y^2\) in Quantity A.
You are told that y ≠ 0, so \(2y^2\) is always positive, and Quantity A will always therefore be larger. The answer is choice (A).
Plugging in 2 for y gives you \(x^2 = 5\) in the given equation and 17 for Quantity B.
Squaring this gives you \(x^4 = 25\) for Quantity A, which is therefore larger.
Plugging in any other number gives the same result.
Alternatively, doing algebra by squaring both sides of the given equation reveals Quantity A: \(x4 = (y^2 + 1)(y^2 + 1) = y^4 + 2y^2 + 1\).
The only difference between Quantities A and B is the \(2y^2\) in Quantity A.
You are told that y ≠ 0, so \(2y^2\) is always positive, and Quantity A will always therefore be larger. The answer is choice (A).
