Gprabhumir wrote:
Hi,
I was practicing the gmat flashcards available on gmatclub.
https://gmatclub.com/forum/gmat-flashcards-108651.htmlOn page 28, if you see the solution, it says sqrt(x-3)^2 is (3-x).
How is that possible
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 \(|\text{some expression}| = -(\text{some expression})\).
When \(x \ge 0\), then \(|x|=x\), or more generally, when \(\text{some expression} \ge 0\), then \(|\text{some expression}| = \text{some expression}\).
Another important property to remember is \(\sqrt{x^2}=|x|\).
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, since the expressions under the square roots are greater than or equal to zero, then \(2-x \ge 0\), which implies \(x \le 2\). Next, since \(x \le 2\), the expression \(x-3\) inside the absolute value is negative. Therefore, \(|x-3|=-(x-3)=-x+3\). This allows us to rewrite the above expression as follows:
\(|x-3|+\sqrt{2-x}+x-3=\)
\(=-x+3+\sqrt{2-x}+x-3=\)
\(=\sqrt{2-x}\).
Answer: A
You can find a detailed discussion of this question here:
https://gmatclub.com/forum/if-each-expr ... 87153.html