M08-16 : GMAT Club Tests

buddyisraelgmat

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Re: M08-16 [#permalink]

20 Jan 2015, 06:49

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Bunuel wrote:

Official Solution:

If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)

Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).

Now, as the expressions under the square roots are more than or equal to zero than \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).

Answer: A

Hi Bunuel

I understand the properties but unable to comprehend the highlighted part.

It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with
\(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\)

Pls help understand this

Thanks

L

Bunuel

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Re: M08-16 [#permalink]

20 Jan 2015, 07:26

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buddyisraelgmat wrote:

Bunuel wrote:

Official Solution:

If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)

Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).

Now, as the expressions under the square roots are more than or equal to zero than \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).

Answer: A

Hi Bunuel

I understand the properties but unable to comprehend the highlighted part.

It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with
\(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\)

Pls help understand this

Thanks

When \(x \le 2\) then x-3 (expression in modulus) is less than 0, thus \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

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Re: M08-16 [#permalink]

08 Aug 2015, 23:47

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buddyisraelgmat wrote:

Bunuel wrote:

Official Solution:

If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)

Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).

Now, as the expressions under the square roots are more than or equal to zero than \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).

Answer: A

Hi Bunuel

I understand the properties but unable to comprehend the highlighted part.

It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with
\(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\)

Pls help understand this

Thanks

Bunuel,

I still not do not understand that when the question stem states that expressions under the root are positive, then why have you considered |x-3| ?
We should then only consider the positive option (x-3)
I chose (x-3) for the first term and ended up as option B

Am i missing anything?

P

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Re: M08-16 [#permalink]

16 Jul 2016, 21:32

I think this is a high-quality question and I agree with explanation. Brilliant Question !

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banerjeeankush99

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Re: M08-16 [#permalink]

26 Sep 2016, 12:42

This is a really great question. Thanks Bunuel :-D

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Re: M08-16 [#permalink]

22 Jan 2017, 05:32

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I think this is a high-quality question and I agree with explanation.

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prateeksaxena55

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Re: M08-16 [#permalink]

02 Jul 2017, 05:44

Bunuel wrote:

Official Solution:

If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)

Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).

Now, as the expressions under the square roots are more than or equal to zero than \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).

Answer: A

Hey Bunuel, i have a doubt, it will be of great help if you can reply. Since the question tells you that the quantity under the root should be positive, why are you considering x-3 to be positive. the quantity under the root is (x-3)^2, which is always positie irrespective of what value x takes.

L

Bunuel

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Re: M08-16 [#permalink]

02 Jul 2017, 21:22

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prateeksaxena55 wrote:

Bunuel wrote:

Official Solution:

If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)

Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).

Now, as the expressions under the square roots are more than or equal to zero than \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).

Answer: A

Hey Bunuel, i have a doubt, it will be of great help if you can reply. Since the question tells you that the quantity under the root should be positive, why are you considering x-3 to be positive. the quantity under the root is (x-3)^2, which is always positie irrespective of what value x takes.

The point is that \(\sqrt{x^2}=|x|\). Next, \(|x| = x\), when \(x \geq 0\) and \(|x| = -x\), when \(x \leq 0\).

We have \(\sqrt{x^2 - 6x + 9}=\sqrt{(x-3)^2}\). Thus according to the above property \(\sqrt{x^2 - 6x + 9}=\sqrt{(x-3)^2}=|x-3|\). Next, since \(x \le 2\), then \(x - 3 < 0\) and therefore \(|x-3| = -(x-3) = 3-x\).

Notice here that since \(x \le 2\), then 3 - x will be positive, so the condition that the square root must give a non-negative result is satisfied.

B

Terabyte

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Re: M08-16 [#permalink]

24 Oct 2017, 23:53

In the GMAT Club math book:
"When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root".
That is, sqrt(25) = 5, NOT +-5.

So, why sqrt(x^2) must be |x|, not just +x?

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Re: M08-16 [#permalink]

25 Oct 2017, 00:00

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Terabyte wrote:

In the GMAT Club math book:
"When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root".
That is, sqrt(25) = 5, NOT +-5.

So, why sqrt(x^2) must be |x|, not just +x?

\(\sqrt{x^2}=|x|\):[/b]

The point here is that since square root function cannot give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

Hope it's clear.

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Re: M08-16 [#permalink]

01 Nov 2017, 08:23

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I think this is a high-quality question and I agree with explanation.

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Shobhit7

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Re: M08-16 [#permalink]

09 Nov 2017, 01:51

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Using the plug-in method is a faster way to get correct answer.

As we know that expression under square root is <=0, x must be <=2.

Plug in any value of x that is <=2, say x=1, in given expressionand each answer choice.

Only Choice A will match.

Posted from my mobile device

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Re: M08-16 [#permalink]

12 Aug 2018, 11:25

What a great question.

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Re: M08-16 [#permalink]

07 Jan 2019, 21:10

Fab Question!

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Re: M08-16 [#permalink]

27 Sep 2019, 23:49

I think this is a high-quality question and I agree with explanation.

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Re: M08-16 [#permalink]

15 Feb 2020, 12:45

I think this is a high-quality question and I agree with explanation.

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Re: M08-16 [#permalink]

26 Apr 2020, 05:14

Good question.

gmatclubot

Re: M08-16 [#permalink]

26 Apr 2020, 05:14