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The equation ax^ 2= bx^2 + 1, where a and b are constants,
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26 Nov 2019, 11:04
26 Nov 2019, 11:04
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The equation \(ax^2 = bx^2 + 1\), where a and b are constants, has two real solutions.
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.
Kudos for the right answer and explanation
Quantity A |
Quantity B |
\(a\) |
\(b\) |
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.
Kudos for the right answer and explanation
Re: The equation ax^ 2= bx^2 + 1, where a and b are constants,
[#permalink]
26 Nov 2019, 19:17
26 Nov 2019, 19:17
5
ax^2 = bx^2 + 1, where a and b are constants, has two real solutions
(a-b)x^2 = 1
For the solutions to exist x^2 >= 0, thus
a-b>0
(cant be equal to zero, because the equation wont hold true)
=> a>b
(A)
(a-b)x^2 = 1
For the solutions to exist x^2 >= 0, thus
a-b>0
(cant be equal to zero, because the equation wont hold true)
=> a>b
(A)
General Discussion
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The equation ax^ 2= bx^2 + 1, where a and b are constants,
[#permalink]
12 Feb 2021, 03:11
12 Feb 2021, 03:11
2
\(ax^2\) = \(bx^2 + 1\)
\((a - b)x^2 = 1\)
\(x\) = \(\sqrt{\frac{1}{(a - b)}}\)
Since, the equation has 2 real solutions; (a - b) cannot be zero [will go infinite] or negative [otherwise it will take a form of complex number]
Therefore, (a - b) > 0
i.e. a > b
Hence, option A
\((a - b)x^2 = 1\)
\(x\) = \(\sqrt{\frac{1}{(a - b)}}\)
Since, the equation has 2 real solutions; (a - b) cannot be zero [will go infinite] or negative [otherwise it will take a form of complex number]
Therefore, (a - b) > 0
i.e. a > b
Hence, option A
