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# M08-16

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Math Expert
Joined: 02 Sep 2009
Posts: 52463

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15 Sep 2014, 23:37
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Difficulty:

85% (hard)

Question Stats:

41% (00:51) correct 59% (00:51) wrong based on 261 sessions

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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. $$x + \sqrt{x - 2}$$

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Joined: 02 Sep 2009
Posts: 52463

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15 Sep 2014, 23:37
7
4
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 then $$2-x \ge 0$$ $$\rightarrow$$ $$x \le 2$$. Next: as $$x \le 2$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

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

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Joined: 14 Jul 2014
Posts: 92

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20 Jan 2015, 05:49
2
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 $$|x-3|=-(x-3)=-x+3$$.

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

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
Math Expert
Joined: 02 Sep 2009
Posts: 52463

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20 Jan 2015, 06:26
6
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 $$|x-3|=-(x-3)=-x+3$$.

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

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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Joined: 14 Jul 2014
Posts: 92

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20 Jan 2015, 06:42
2
Bunuel wrote:
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 $$|x-3|=-(x-3)=-x+3$$.

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

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 $$|x-3|=-(x-3)=-x+3$$.

Oh Ok. Logically makes sense. Thanks
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Location: India
Concentration: Social Entrepreneurship, General Management
WE: Engineering (Investment Banking)

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08 Aug 2015, 22:47
1
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 $$|x-3|=-(x-3)=-x+3$$.

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

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?
_________________

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Joined: 30 Jul 2014
Posts: 11
Location: Brazil
Concentration: Other, Strategy
GPA: 3.95
WE: Consulting (Education)

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14 Aug 2015, 11:57
2
1
sagar2911 wrote:
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 $$|x-3|=-(x-3)=-x+3$$.

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

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?

Because the question says "each expression under the square root is greater than or equal to 0", we need to understand that if x-3 is 0, then 2-x is a negative number, so the root of 2 - x is root of -1 what breaks the rule stated in the question.
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Joined: 08 Sep 2012
Posts: 61
Location: India
Concentration: Social Entrepreneurship, General Management
WE: Engineering (Investment Banking)

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14 Aug 2015, 23:33
1
roneileonel wrote:
sagar2911 wrote:

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?

Because the question says "each expression under the square root is greater than or equal to 0", we need to understand that if x-3 is 0, then 2-x is a negative number, so the root of 2 - x is root of -1 what breaks the rule stated in the question.

Thank you dude! This helps a lot!
_________________

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GMAT Date: 12-06-2014

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11 Feb 2016, 08:41
schak2rhyme wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Next: as x≤2 then |x−3| becomes |x−3|=−(x−3)=−x+3.

In the excerpt from the solution above why is |x−3| not >=0 instead of <=.Please explain.

The question says the expression cant be negative .. not roots .. x=-3 because the whole expression root of 2-x should be positive ..even if x is negative the first expression remains positive .. am i correct ?
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Joined: 21 Sep 2015
Posts: 77
Location: India
GMAT 1: 730 Q48 V42
GMAT 2: 750 Q50 V41

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16 Jul 2016, 20:32
I think this is a high-quality question and I agree with explanation. Brilliant Question !
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Appreciate any KUDOS given !

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01 Sep 2016, 20:09
Please let me know if I did this correctly alternatively.

$$\sqrt{x^2 - 6x + 9}$$ , solving (x-3)*(x-3) , x = 3,

plug in 3 for all x, becomes answer A
Math Expert
Joined: 02 Sep 2009
Posts: 52463

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02 Sep 2016, 02:58
xnthic wrote:
Please let me know if I did this correctly alternatively.

$$\sqrt{x^2 - 6x + 9}$$ , solving (x-3)*(x-3) , x = 3,

plug in 3 for all x, becomes answer A

No, that's not correct. Why are you equating $$\sqrt{x^2 - 6x + 9}$$ to 0? Why should x be 3? As shown in the solution x must be less than or equal to 2 for $$\sqrt{2-x}$$ to be defined.
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26 Sep 2016, 11:42
This is a really great question. Thanks Bunuel
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22 Jan 2017, 04:32
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 12 Jan 2017
Posts: 1

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02 Jul 2017, 04:44
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 $$|x-3|=-(x-3)=-x+3$$.

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

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.
Math Expert
Joined: 02 Sep 2009
Posts: 52463

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02 Jul 2017, 20:22
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 $$|x-3|=-(x-3)=-x+3$$.

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

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.
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04 Jul 2017, 20:48
The question says " each expression under the square root is greater than or equal to 0"
Hence, It comes out that x is less than or equal to 2 by second expression and for the first expression why at all we are thinking it to be square of (x-3) as this will result in values greater or equal to 3 not valid as it is less than 2. but if we take it as square of (3-x) then it will less than 3 satisfying outcome from other equation.

Therefore, the shorter method of solving will be as follows.

Step 1. from second expression find x is less than or equal to 2
Step 2. solve first expression as sqaure of (3-x) not square of (x-3) { as this gives x is greater or equal to 3} and then you get the answer.

It avoids complexity of modulus.
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Posts: 106

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24 Oct 2017, 22:53
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?
Math Expert
Joined: 02 Sep 2009
Posts: 52463

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24 Oct 2017, 23:00
2
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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24 Oct 2017, 23:05
Bunuel wrote:

$$\sqrt{x^2}=|x|$$:[/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.

Yes, it's clear. Thank you very much, Bunuel. Learning a lot from your posts. Aiming for Q50
Re: M08-16 &nbs [#permalink] 24 Oct 2017, 23:05

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# M08-16

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