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Intern  B
Joined: 04 May 2017
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I think this is a high-quality question and I agree with explanation.
Manager  G
Joined: 01 Feb 2017
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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 expression​and each answer choice.

Only Choice A will match.

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Intern  B
Joined: 24 Jun 2013
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Bunuel wrote:
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. $$x + \sqrt{x - 2}$$

Hie Bunuel,

Math Expert V
Joined: 02 Sep 2009
Posts: 55801

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prk0789 wrote:
Bunuel wrote:
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. $$x + \sqrt{x - 2}$$

Hie Bunuel,

8. Exponents and Roots of Numbers

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I think this is a high-quality question and I agree with explanation.
Manager  G
Joined: 26 Dec 2017
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Here as per condition we can get both x<=2 and x>=3
wondering how only x<=2 condition was considered initially
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Math Expert V
Joined: 02 Sep 2009
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tejyr wrote:
Here as per condition we can get both x<=2 and x>=3
wondering how only x<=2 condition was considered initially

The square of a number is 0 or positive, so from $$\sqrt{(x-3)^2}$$ we don't have x >= 3 because (x - 3)^2 is non-negative for any value of x, not only for x >= 3.
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Manager  S
Joined: 26 Feb 2018
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GMAT 1: 640 Q45 V34 GPA: 3.9
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Hey Bunuel!

I'm a bit confused over here. I do understand that $$\sqrt{x^2} = |x|$$. However, we are given that each expression under the square root is greater than or equal to 0. Hence $$\sqrt{x^2−6x+9} = \sqrt{(x-3)^2} = |x-3|$$.

So the entity, $$|x-3| ≥ 0$$ Hence, isn't the range of x becomes $$x≥3$$ from this term?

Let me know where am I going wrong.

Thanks a lot
PrayasT

Originally posted by PrayasT on 22 Jun 2018, 14:46.
Last edited by PrayasT on 22 Jun 2018, 23:02, edited 1 time in total.
Math Expert V
Joined: 02 Sep 2009
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PrayasT wrote:
Hey Bunnel!

I'm a bit confused over here. I do understand that $$\sqrt{x^2} = |x|$$. However, we are given that each expression under the square root is greater than or equal to 0. Hence $$\sqrt{x^2−6x+9} = \sqrt{(x-3)^2} = |x-3|$$.

So the entity, $$|x-3| ≥ 0$$ Hence, isn't the range of x becomes $$x≥3$$ from this term?

Let me know where am I going wrong.

Thanks a lot
PrayasT

Absolute value is always non-negative so $$|x-3| ≥ 0$$ is true for any value of x. Or in other way, since the square of a number is non-negative then $$\sqrt{(x-3)^2}$$ is defined for any value of x.

We get the range of x from $$\sqrt{2 - x}$$. The square root to be defined the expression under the square root must be greater than or equal to 0 --> $$2-x \ge 0$$ $$\rightarrow$$ $$x \le 2$$.
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Manager  S
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GMAT 1: 640 Q45 V34 GPA: 3.9
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Bunuel wrote:
PrayasT wrote:
Hey Bunnel!

I'm a bit confused over here. I do understand that $$\sqrt{x^2} = |x|$$. However, we are given that each expression under the square root is greater than or equal to 0. Hence $$\sqrt{x^2−6x+9} = \sqrt{(x-3)^2} = |x-3|$$.

So the entity, $$|x-3| ≥ 0$$ Hence, isn't the range of x becomes $$x≥3$$ from this term?

Let me know where am I going wrong.

Thanks a lot
PrayasT

Absolute value is always non-negative so $$|x-3| ≥ 0$$ is true for any value of x. Or in other way, since the square of a number is non-negative then $$\sqrt{(x-3)^2}$$ is defined for any value of x.

We get the range of x from $$\sqrt{2 - x}$$. The square root to be defined the expression under the square root must be greater than or equal to 0 --> $$2-x \ge 0$$ $$\rightarrow$$ $$x \le 2$$.

That makes perfect sense! Thanks a lot!

Regards,
PrayasT
Intern  B
Joined: 08 Aug 2017
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I solved in the following manner:

Once you realize x<=2, plug in x=2 in the prompt, you get 0.
Now check answer choices by plugging in x=2 to get the value 0.

Intern  B
Joined: 26 Dec 2016
Posts: 34
Concentration: Finance, Entrepreneurship
GMAT 1: 710 Q45 V41 GMAT 2: 610 Q41 V33 GMAT 3: 710 Q47 V40 GPA: 3.5

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What a great question.
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Intern  B
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Hi Bunuel,

You have made a point that:
"When the GMAT provides the square root sign for an even root, such as a square root, fourth root, erc. then the only accepted answer is the positive root.

That is:
√9=3, NOT +3 or -3;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3."

Why same is not valid here?
Why are we considering the value of sqrt as positive and negative?
Math Expert V
Joined: 02 Sep 2009
Posts: 55801

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SSwami wrote:
Hi Bunuel,

You have made a point that:
"When the GMAT provides the square root sign for an even root, such as a square root, fourth root, erc. then the only accepted answer is the positive root.

That is:
√9=3, NOT +3 or -3;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3."

Why same is not valid here?
Why are we considering the value of sqrt as positive and negative?

Where do you see a contradiction? When we have that $$\sqrt{x^2}=|x|$$, still the square root is giving a non-negative result because the absolute value of a number is always non-negative. Please read the previous pages, I think several posts there explain this clearly.
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Intern  B
Joined: 18 Dec 2018
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Why is the expression >=0 but becomes <=0 in the explanation? If according to the setting of the question, shouldn't the answer be D?
Intern  B
Joined: 18 Dec 2018
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Because we know X<=2, -(X-3) instead of (X-3).
But how do we make sure if -(X-3) satisfy X<=2?
Intern  B
Joined: 22 Oct 2018
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GMAT 1: 770 Q50 V45 Show Tags

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