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Is x>0? 1) x^2=9x 2) x^2=81

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Is x>0? 1) x^2=9x 2) x^2=81  [#permalink]

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New post Updated on: 02 Feb 2016, 16:45
1
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A
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C
D
E

Difficulty:

  45% (medium)

Question Stats:

53% (00:58) correct 47% (00:51) wrong based on 234 sessions

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Is x>0?

1) x^2=9x
2) x^2=81

Originally posted by Abrax99 on 02 Feb 2016, 16:39.
Last edited by ENGRTOMBA2018 on 02 Feb 2016, 16:45, edited 1 time in total.
Reformatted the question and edited the title
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Re: Is x>0? 1) x^2=9x 2) x^2=81  [#permalink]

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New post 02 Feb 2016, 16:53
2
1
Abrax99 wrote:
Is x>0?

1) x² = 9x
2) x² = 81


Target question: Is x>0?

Statement 1: x² = 9x
Rearrange to get: x² - 9x = 0
Factor to get x(x - 9) = 0
So, we have two possible values for x.
Case a: x = 0, in which case x is NOT greater than 0
Case b: x = 9, in which case x IS greater than 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x² = 81
Rearrange to get: x² - 81 = 0
Factor to get (x + 9)(x - 9) = 0
So, we have two possible values for x.
Case a: x = -9, in which case x is NOT greater than 0
Case b: x = 9, in which case x IS greater than 0
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that x = 0 or 9
Statement 2 tells us that x = -9 or 9
Since only one value of x satisfies BOTH statements, it MUST be the case that x = 9, which means x IS greater than 0
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent
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Re: Is x>0? 1) x^2=9x 2) x^2=81  [#permalink]

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New post 27 Feb 2017, 11:35
Hi Brent,

My approach was as follows:

x squared/x =9x/x
Therefore, x=9
So A is sufficient
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Re: Is x>0? 1) x^2=9x 2) x^2=81  [#permalink]

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New post 27 Feb 2017, 12:08
1
TayoUmar wrote:
Hi Brent,

My approach was as follows:

x squared/x =9x/x
Therefore, x=9
So A is sufficient

Hi TayoUmar,
This way is a bit faulty, as it forgets one solution x=0, also dividing by zero is a bit not so right. But the bottomline is x could be zero.

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Re: Is x>0? 1) x^2=9x 2) x^2=81  [#permalink]

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New post 27 Feb 2017, 15:12
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Top Contributor
TayoUmar wrote:
Hi Brent,

My approach was as follows:

x squared/x =9x/x
Therefore, x=9
So A is sufficient


We should always be wary of dividing both sides of an equation by a variable, since this could yield unintended consequences.
Consider this example equation: x² = xy
If we divide both sides by x it SEEMS like we get an equivalent equation: x = y

However, if we replace the variables with actual numbers, we can see where the problem arises.
Take x² = xy and plug in 0 for x and 3 for y.
We: (0)(0) = (0)(3) [we can all agree on the validity of this equation]
Since there's a 0 on both sides, let's divide both sides by 0 to get: 0 = 3 [hmmmmm]

So, when we took x² = xy and divided both sides by x, we never considered the possibility that x might equal 0.

Here's a useful way to solve x² = xy
Subtract xy from both sides to get: x² - xy = 0
Factor: x(x - y) = 0
This means either x = 0, or x-y = 0

Cheers,
Brent
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Re: Is x>0? 1) x^2=9x 2) x^2=81  [#permalink]

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New post 16 Jul 2017, 11:52
Is x>0?

1) \(x^2=9x\)

\(x^2=9x\)

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

\(x(x-9) = 0\)

\(x = 0\)

OR

\(x = 9\)

Hence, (1) ===== is NOT SUFFICIENT

2) \(x^2=81\)

\(x^2=81\)

\(x = +9\) OR

\(x = -9\)

Hence, (2) ===== is NOT SUFFICIENT

Combining (1) & (2)

We get \(x = 9\)

Hence, (1) & (2) ===== is SUFFICIENT

Hence, Answer is C

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Re: Is x>0? 1) x^2=9x 2) x^2=81  [#permalink]

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New post 03 Apr 2018, 16:53
Abrax99 wrote:
Is x>0?

1) x^2=9x
2) x^2=81


1) x^2=9x

x^2- 9x = 0

x (x - 9) = 0

x = 0 or x = 9

No certain answer

Insufficient

2) x^2=81

x =9 or x =-9

No certain answer

Insufficient

Combine 1 & 2

x = 9

Answer is always yes

Answer: C
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Re: Is x>0? 1) x^2=9x 2) x^2=81 &nbs [#permalink] 03 Apr 2018, 16:53
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