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# M08-16

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Joined: 04 May 2017
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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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09 Nov 2017, 01:51
2
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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23 Nov 2017, 04:28
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,

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23 Nov 2017, 05:28
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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11 Feb 2018, 14:14
1
I think this is a high-quality question and I agree with explanation. There is a grammar error on the explanation. Instead of "than" there should be "then" before ":"
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11 Feb 2018, 21:49
rodmm wrote:
I think this is a high-quality question and I agree with explanation. There is a grammar error on the explanation. Instead of "than" there should be "then" before ":"

Edited the typo. Thank you.
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20 Feb 2018, 02:00
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I think this is a high-quality question and I agree with explanation.
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02 Jun 2018, 06:27
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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02 Jun 2018, 09:27
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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Updated on: 22 Jun 2018, 23:02
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.
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22 Jun 2018, 22:34
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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22 Jun 2018, 23:05
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
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Joined: 08 Aug 2017
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20 Jul 2018, 21:36
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.

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12 Aug 2018, 11:25
What a great question.
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Re: M08-16 &nbs [#permalink] 12 Aug 2018, 11:25

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# M08-16

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