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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 30 Jun 2018, 09:09
nigina93 wrote:
Hi Bunuel, chetan2u, VeritasPrepKarishma
I used the following approach, not sure if its the right reasoning though, can you please check it out
x^2-6x+9=0
D=(6^2)36-4*9=0
x=(6+-0)=3,
so x is 3. substitute x with 0 in the expression and we get A as correct answer only.


No, this is not right at all.

We don't have an equation here. We have an expression. From where did you get that x^2 - 6x + 9 = 0? And if x = 3, why are you substituting x = 0, then?
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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, 00:47
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, 01:01
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, 07:53
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, 09:07
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, 20:32
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]

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New post 02 Apr 2020, 09:01
Hi Bunuel,

I understand from your explanation is that \sqrt{x^2} = |x| . So according to this \sqrt{x^2-6x-9} => \sqrt{(x-3)^2} => |x-3|

My doubt is : In question it is mention that under the square root is greater than or equal to 0 so as |x-3| also comes out from the square root hence it should be positive only but you have taken it as negative due to x<=2 condition.

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

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New post 02 Apr 2020, 09:09
a123bansal wrote:
Hi Bunuel,

I understand from your explanation is that \sqrt{x^2} = |x| . So according to this \sqrt{x^2-6x-9} => \sqrt{(x-3)^2} => |x-3|

My doubt is : In question it is mention that under the square root is greater than or equal to 0 so as |x-3| also comes out from the square root hence it should be positive only but you have taken it as negative due to x<=2 condition.

Please Help.


An absolute value is always more than or equal to zero. So, |x - 3| >=0. Next, when x <= 2, then x - 3 < 0, and thus |x - 3| = -(x - 3), but -(x - 3) is POSITIVE when x <= 2. So, nothing is violated there.

One should brush up basics before attempting questions (especially harder ones).

10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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Re: If each expression under the square root is greater than or   [#permalink] 02 Apr 2020, 09:09

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