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(x^2+9)(x+9)(x-5)=0

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(x^2+9)(x+9)(x-5)=0 [#permalink] New post   19 Mar 2021, 06:39
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\((x^2+9)(x+9)(x-5)=0\)


Quantity A
Quantity B
x
7


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
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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(x^2+9)(x+9)(x-5)=0 [#permalink] New post   19 Mar 2021, 07:37
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\((x^2+9)(x+9)(x-5)\)=0

\((x^2+9)\) can never be 0 so (x+9) or (x-5) or both has to be zero.

For the equation to be zero x will be 5 OR -9

Either value of x is less than 7.

Qt A < Qt B

Answer B
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(x^2+9)(x+9)(x-5)=0 [#permalink] New post   19 Mar 2021, 07:58
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Carcass wrote:
\((x^2+9)(x+9)(x-5)=0\)

Quantity A
Quantity B
x
7


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.


\((x^2+9)(x+9)(x-5)=0\)

[ \(x^2 + 9\) can never be a zero, because \(x^2 ≠ -9\) ]

Since, we can have either \((x+9) = 0\) or \((x-5) = 0\)
i.e. \(x = -9\) or \(x = 5\)

\(7\) would always be greater than \(5\) or \(-9\)

Hence, option B
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Re: (x^2+9)(x+9)(x-5)=0 [#permalink] New post   24 Mar 2021, 03:09
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The basic idea in this questions is that A x B =0 only when A=0 or B=0 or both=0

Similarly, in the given question, one of the terms has to be 0,
thus we have a total of 3 possible equations.

\(x^2+9=0\) implying that x is an imaginary number since square of any real number is always positive
\(x+9=0\) implying that x can be -9
\(x-5=0\) implying that x can be 5

Both the real possible values of x are lesser than part B of the question, hence Option B is correct because 7 will always be greater than -9 or 5
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Re: (x^2+9)(x+9)(x-5)=0 [#permalink] New post   16 Aug 2024, 09:19
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Re: (x^2+9)(x+9)(x-5)=0 [#permalink]
16 Aug 2024, 09:19
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