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The equation ax^ 2= bx^2 + 1, where a and b are constants,

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The equation ax^ 2= bx^2 + 1, where a and b are constants, [#permalink] New post   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.


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.

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Re: The equation ax^ 2= bx^2 + 1, where a and b are constants, [#permalink] New post   26 Nov 2019, 19:17
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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)
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The equation ax^ 2= bx^2 + 1, where a and b are constants, [#permalink] New post   12 Feb 2021, 03:11
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\(ax^2\) = \(bx^2 + 1\)

\((a - b)x^2 = 1\)

\(x\) = \(\sqrt{\frac{1}{(a - b)}}\)

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
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Re: The equation ax^ 2= bx^2 + 1, where a and b are constants, [#permalink] New post   13 Feb 2021, 11:12
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x can't be zero since the two sides of the equation wouldn't be equal if it was; thus we can safely divide both sides of the equation by x^2, which leaves us with a = b + (1/x^2). In other words, a equals b plus a positive number (since x^2 will always be positive). Thus, a will always be greater than b.

(A) is the correct answer.
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Re: The equation ax^ 2= bx^2 + 1, where a and b are constants, [#permalink] New post   10 Aug 2022, 00:31
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Re: The equation ax^ 2= bx^2 + 1, where a and b are constants, [#permalink]
10 Aug 2022, 00:31
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