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x^2=y^2+1
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Updated on: 17 Sep 2019, 09:24
Updated on: 17 Sep 2019, 09:24
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Question Stats:
79% (01:14) correct
20% (04:18) wrong based on 24 sessions
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Q- ID: Q02-16
\(x^2=y^2+1\) and \(y \neq 0\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
I tried to solve this by plugin numbers but when I tried to take x=1/2
y^2=-3/4 which doesn't exist.
May someone clarify pls?
\(x^2=y^2+1\) and \(y \neq 0\)
Quantity A |
Quantity B |
\(x^4\) |
\(-y^4+1\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
I tried to solve this by plugin numbers but when I tried to take x=1/2
y^2=-3/4 which doesn't exist.
May someone clarify pls?
Re: x^2=y^2+1
[#permalink]
17 Sep 2019, 10:49
17 Sep 2019, 10:49
Expert Reply
QA
\((x^2)^2\)
\(x^4 = (y^2+1)^2\)
QB
\(y^2=x^2-1\)
\(-(y^2)^2 +1\)
\(-(x^2-1)^2+1\)
\(-(x^4+1-2x^2)+1\)
\(-x^4-1+2x^2+1\)
\(x^4-2x^2\)
\(x^2(x^2-2)=0\)
\(x^2=0\) and \(x^2=2\)
\(x^2=0\) and \(x=\sqrt{2}\)
Qa will be always > of QB
Regards
\((x^2)^2\)
\(x^4 = (y^2+1)^2\)
QB
\(y^2=x^2-1\)
\(-(y^2)^2 +1\)
\(-(x^2-1)^2+1\)
\(-(x^4+1-2x^2)+1\)
\(-x^4-1+2x^2+1\)
\(x^4-2x^2\)
\(x^2(x^2-2)=0\)
\(x^2=0\) and \(x^2=2\)
\(x^2=0\) and \(x=\sqrt{2}\)
Qa will be always > of QB
Regards
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x^2=y^2+1
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17 Sep 2019, 11:05
17 Sep 2019, 11:05
Asmakan wrote:
Q- ID: Q02-16
\(x^2=y^2+1\) and \(y \neq 0\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
\(x^2=y^2+1\) and \(y \neq 0\)
Quantity A |
Quantity B |
\(x^4\) |
\(-y^4+1\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Here's another approach:
Given: \(x^2=y^2+1\)
Square both sides to get: \((x^2)^2=(y^2+1)^2\)
Simplify both sides to get: \(x^4=y^4+2y^2+1\)
We'll use this information shortly
-------------------------------------
We'll start with:
Quantity A: \(x^4\)
Quantity B: \(-y^4+1\)
Replace \(x^4\) with \(y^4+2y^2+1\) to get:
Quantity A: \(y^4+2y^2+1\)
Quantity B: \(-y^4+1\)
Subtract 1 from both quantities to get:
Quantity A: \(y^4+2y^2\)
Quantity B: \(-y^4\)
Subtract add \(y^4\) to both quantities to get:
Quantity A: \(2y^4+2y^2\)
Quantity B: \(0\)
Factor Quantity A to get:
Quantity A: \(2(y^4+y^2)\)
Quantity B: \(0\)
Since \(y \neq 0\), we know that \(y^4\) is POSITIVE and \(y^2\) is POSITIVE
So, \(y^4+y^2\) is POSITIVE
We get:
Quantity A: 2(some positive number)
Quantity B: 0
Clearly Quantity A is greater.
Answer: A
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