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

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Joined: 23 Feb 2015
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Re: If each expression under the square root is greater than or  [#permalink]

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27 May 2019, 01:47
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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Posts: 59712
Re: If each expression under the square root is greater than or  [#permalink]

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27 May 2019, 01:52
1
1
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]

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27 May 2019, 02:01
Bunuel wrote:
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 $$|x-3|=-(x-3)=-x+3$$.

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

Oh, I forgot to change the answer option with my changing version. Thanks anyway.
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Re: If each expression under the square root is greater than or  [#permalink]

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27 May 2019, 08:53
But on Gmat only real values are considered. But if √x^2 is negative then its a complex number . And in Gmat complex numbers are not used

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

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31 Aug 2019, 10:07
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) $$x + \sqrt{x-2}$$

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

M18-16

Given: Each expression under the square root is greater than or equal to 0.

Asked: What is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

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

Since all expressions inside square root are non-negative numbers

x^2 - 6x +9 = (x -3)^2 >=0

2- x >=0
x<=2

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

Since x <=2 => x -3 <= -1 <0 => |x-3| = 3-x

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

IMO A
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Re: If each expression under the square root is greater than or  [#permalink]

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25 Nov 2019, 21:32
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) $$x + \sqrt{x-2}$$

Quick Solution here
We know that the stem mentions that the value under the square roots have to equal 0 or larger, so you need to use a value that's less than 2.
Let's say X = 1, when you plug it in the stem, you get that expression to equal 1.

Only Answer Choice A matches that value
Re: If each expression under the square root is greater than or   [#permalink] 25 Nov 2019, 21:32

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