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If each expression under the square root is greater than or

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

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

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New post 27 May 2019, 01:52
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Asad 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 \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{3-x}+x-3=-x+3+\sqrt{3-x}+x-3=\sqrt{3-x}\).
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Re: If each expression under the square root is greater than or  [#permalink]

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New post 27 May 2019, 02:01
Bunuel wrote:
Asad 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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New post 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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New post 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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New post 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
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Re: If each expression under the square root is greater than or   [#permalink] 25 Nov 2019, 21:32

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