M08-16

Bunuel,

I still not do not understand that when the question stem states that expressions under the root are positive, then why have you considered |x-3| ?
We should then only consider the positive option (x-3)
I chose (x-3) for the first term and ended up as option B

Am i missing anything?

Official Solution:

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. \(x + \sqrt{x - 2}\)


Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

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

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

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


Answer: A

Hi Bunuel

I understand the properties but unable to comprehend the highlighted part.

It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with
\(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\)

Pls help understand this

Thanks

Hi Bunuel

I understand the properties but unable to comprehend the highlighted part.

It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with
\(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\)

Pls help understand this

Thanks

Official Solution:

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. \(x + \sqrt{x - 2}\)


Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

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

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

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


Answer: A

Hey Bunuel, i have a doubt, it will be of great help if you can reply. Since the question tells you that the quantity under the root should be positive, why are you considering x-3 to be positive. the quantity under the root is (x-3)^2, which is always positie irrespective of what value x takes.

In the GMAT Club math book:
"When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root".
That is, sqrt(25) = 5, NOT +-5.

So, why sqrt(x^2) must be |x|, not just +x?