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# If each expression under the square root is greater than or

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Q50  V47
Re: If each expression under the square root is greater than or [#permalink]
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study

Quote:
Even if x <= 2, rt (x-3)^2 is still a valid explanation. Since there is a square below the rt sign the number is obviously positive. How can one conveniently change from (x - 3) to (3 - x)?

yes, you can and in fact should change it to (3-x). Here's explanation:

https://en.wikipedia.org/wiki/Square_root

$$sqrt(x^2) = |x|$$

|x| = x, for all x>=0
|x| = -x, for all x<0

In our case we have $$\sqrt{(x-3)^2}=|x-3|=-(x-3)=3-x$$, because (2-x)>=0; x<=2; (x-3)<0

And so the correct answer is

$$3-x+\sqrt{2-x}+x-3=\sqrt{2-x}$$
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Re: If each expression under the square root is greater than or [#permalink]
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study wrote:
Can someone please explain the logic here. This problem is from M08.

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

* $$\sqrt{2 - x}$$
* $$2x - 6 + \sqrt{2 - x}$$
* $$\sqrt{2 - x} + {x - 3}$$
* $$2x - 6 + \sqrt{x - 2}$$
* $$x + \sqrt{x - 2}$$

First, $$(2 - x)$$ has to be $$\ge 0$$ for the expression $$\sqrt{2 - x}$$ to be defined (GMAT deals only with real numbers). Therefore, $$x \le 2$$ .

Even if x <= 2, rt (x-3)^2 is still a valid explanation. Since there is a square below the rt sign the number is obviously positive. How can one conveniently change from (x - 3) to (3 - x)?

$$\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = 3 - x$$ (because $$x \le 2$$ and thus $$x - 3 \lt 0$$ ).

IMO:
$$\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = x - 3$$ is valid

Summing up: $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + (x - 3) = (x - 3) + \sqrt{2 - x} + (x - 3) = 2x - 6 + \sqrt{x - 2}$$.

What's the OA? I chose (B) I'm afraid I might be wrong though
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Re: If each expression under the square root is greater than or [#permalink]
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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) $$x + \sqrt{x-2}$$

One important note: $$\sqrt{x^2}=|x|$$

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

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

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

I don't follow the highlighted part at all. So if you have x-n, and it is supposed to be positive you can just change around the order? Shouldn't the expression just be undefined, since the facts we are given, and the expression we are given can't coexist?
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Re: If each expression under the square root is greater than or [#permalink]
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AccipiterQ 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) $$x + \sqrt{x-2}$$

One important note: $$\sqrt{x^2}=|x|$$

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

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

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

I don't follow the highlighted part at all. So if you have x-n, and it is supposed to be positive you can just change around the order? Shouldn't the expression just be undefined, since the facts we are given, and the expression we are given can't coexist?

Absolutely not! Ordinarily, if you change the order, it changes everything. But this situation is not ordinary. It is a mod situation i.e. there is modulus around x - n.

Recall how mods are defined:
|x| = x if x is positive
= -x if x is negative

So if you want to get rid of the mod sign, you need to know whether x is positive or negative.

Here we know that x <= 2. For all such values of x, (x-3) is negative.
So |x-3| = -(x - 3) = 3 - x
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Re: If each expression under the square root is greater than or [#permalink]
WOW! That is a really well concealed trap.
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Re: If each expression under the square root is greater than or [#permalink]
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If we just plug in the number 1 we can get the answer in 30 seconds max.
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Re: If each expression under the square root is greater than or [#permalink]
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study 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) $$x + \sqrt{x-2}$$

Main Idea: We are likely looking for cancellation of terms

Details: We see the first term is sqrt(x-3)^2 and looks like the right term for cancellation with (x-3)

sqrt(x-3)^2 = |x-3| which can be x-3 or -(x-3) = -x+3

We can rule out x-3 because x< 2 and so it is negative

We now have the expressions as : -X+3 + sqrt(2-x) +x-3 = sqrt(2-x)

Hence A
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Re: If each expression under the square root is greater than or [#permalink]
study 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) $$x + \sqrt{x-2}$$

Hi Bunuel, IanStewart
If the question stem is changed with below, the correct choice still be A, right?

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

(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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Re: If each expression under the square root is greater than or [#permalink]
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study 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) $$x + \sqrt{x-2}$$

Hi Bunuel, IanStewart
If the question stem is changed with below, the correct choice still be A, right?

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

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

In this case the correct answer would be$$\sqrt{3-x}$$.

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

Now, as the expressions under the square roots are more than or equal to zero than $$3-x\geq{0}$$ --> $$x\leq{3}$$. Next: as $$x\leq{3}$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

$$|x-3|+\sqrt{3-x}+x-3=-x+3+\sqrt{3-x}+x-3=\sqrt{3-x}$$.
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Re: If each expression under the square root is greater than or [#permalink]
Hi KarishmaB, Bunuel
i assumed x=1 and compared the answers with what I got and looks like only A is matching. Is this approach alright or did I just get lucky? My solution:

$$\sqrt{x^{2}-6x+9}+ \sqrt{2-x} + x -3$$

taking x=1

$$\sqrt{1-6(1)+9}+ \sqrt{2-1} + 1 -3$$

$$\sqrt{4}+ \sqrt{1} + -2$$

since it says in the question that everything under the sq root is positive I've taken the positive numbers only:

2+1 -2 = 1

and after solving the answer choices, only A yields the same answer of 1
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Re: If each expression under the square root is greater than or [#permalink]
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ritzu

ritzu wrote:
Hi KarishmaB, Bunuel
i assumed x=1 and compared the answers with what I got and looks like only A is matching. Is this approach alright or did I just get lucky? My solution:

$$\sqrt{x^{2}-6x+9}+ \sqrt{2-x} + x -3$$

taking x=1

$$\sqrt{1-6(1)+9}+ \sqrt{2-1} + 1 -3$$

$$\sqrt{4}+ \sqrt{1} + -2$$

since it says in the question that everything under the sq root is positive I've taken the positive numbers only:

2+1 -2 = 1

and after solving the answer choices, only A yields the same answer of 1

Yes, you are right in your approach.
When we have a variable in the answer options, often plugging in numbers helps make the question very simple.

I need an expression that is equivalent to $$\sqrt{x^2-6x+9}+\sqrt{2-x} + x - 3$$

If it is equivalent to this, then both, this expression and the correct option will give the same result for each valid value of x.

I will put x = 0 since it is acceptable for the given expression (gives no negative square roots)
$$\sqrt{x^2-6x+9}+\sqrt{2-x} + x - 3 = \sqrt{2}$$

Then putting x = 0 in option (A), I get the same result but not in any other option.
Had I got sqrt(2) from another option too, I would have tried putting in x = 1.

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Re: If each expression under the square root is greater than or [#permalink]
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study 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) $$x + \sqrt{x-2}$$

The goal here is to determine which expression is equivalent to $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$.

In this case, we can easily plug in values of x to test for equivalency.
Now let's give ourselves up to 20 seconds to identify a faster approach.
In this case, we can also apply some algebraic factoring techniques to simplify the given expression.
Since the algebraic approach seems both tricky and time consuming, I'm going to test for equivalency.

Key concept: If two expressions are equivalent, they must evaluate to the same value for every possible value of x.
For example, since the expression 2x + 3x is equivalent to the expression 5x, the two expressions will evaluate to the same number for every value of x.
So, if x = 7, the expression 2x + 3x = 2(7) + 3(7) = 14 + 21 = 35, and the expression 5x = 5(7) = 35

Let's evaluate the given expression for $$x = 0$$.

We get: $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3 = \sqrt{0^2 - 6(0) + 9} + \sqrt{2 - 0} + 0 - 3$$
$$= \sqrt{9} + \sqrt{2} - 3$$
$$= 3 + \sqrt{2} - 3$$
$$= \sqrt{2}$$
So, when $$x = 0$$, the value of the given expression is $$\sqrt{2}$$

We'll now evaluate each answer choice for $$x = 0$$ and eliminate those that don't evaluate to $$\sqrt{2}$$

(A) $$\sqrt{2-x}=\sqrt{2-0}=\sqrt{2}$$ KEEP

(B) $$2x-6 + \sqrt{2-x} = 2(0) -6 + \sqrt{2-0} = -6 + \sqrt{2}$$. ELIMINATE

(C) $$\sqrt{2-x} + x-3=\sqrt{2-0} + 0-3=\sqrt{2} -3$$. ELIMINATE

(D) $$2x-6 + \sqrt{x-2} = 2(0) -6 + \sqrt{0-2} = -6 + \sqrt{-2} =$$ no defined value. ELIMINATE

(E) $$x + \sqrt{x-2}= 0 + \sqrt{0-2} = \sqrt{-2} =$$ no defined value. ELIMINATE

The process of elimination, the correct answer is A.
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