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Value of expression

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Value of expression

by krishna kumar » Sun Dec 12, 2010 9:12 am
If each expression under the square root is greater than or equal to zero what is root(x^2 - 6x + 9) + root(2 -x) + x -3 ?


A. root(2 -x)

B. 2x - 6 + root(2 -x)

C. root(2 - x) + x - 3

D. 2x - 6 + root(x - 2)

E. x + root(x - 2)

[spoiler]OA A

[/spoiler]

My answer is B.

Could anyone enlighten me.

Thanks in advance.
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by shovan85 » Sun Dec 12, 2010 10:18 am
krishna kumar wrote:If each expression under the square root is greater than or equal to zero what is root(x^2 - 6x + 9) + root(2 -x) + x -3 ?
All the expression under square root >= 0

When 2 - x >= 0

=> 2 >= x

Thus x is less than equal to 2

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

=> (x - 3) ^ 2 >= 0

=> + (x - 3) >= 0 Or - (x - 3) >= 0 (As square root of x = +/- x)

=> x >= 3 Or x <= 3

As we know x is less than equal to 2 we have to consider -(x - 3) as the root of expression (x^2 - 6x + 9) .

root(x^2 - 6x + 9) + root(2 -x) + x -3

= - (x - 3) + root(2 -x) + x -3

= root(2 -x)

I myself got to B then I solved this in the manner I wanted to get A. I know its quite impossible to visualize this problem in the way I have shown but I hope it is correct.
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by Rahul@gurome » Sun Dec 12, 2010 11:42 am
krishna kumar wrote:If each expression under the square root is greater than or equal to zero what is root(x^2 - 6x + 9) + root(2 -x) + x -3 ?
Given expression: √(x² - 6x + 9) + √(2 - x) + (x - 3) = √(x - 3)² + √(2 - x) + (x - 3)

Now the expressions under square roots are (x - 3)² and (2 - x). Note that (x - 3)² is always greater than or equal to zero. According to question (2 - x) should be greater than or equal to zero too.

(2 - x) ≥ 0 => x ≤ 2

Now note that: √(x - 3)² = |x - 3| = -(x - 3)
This is because (x - 3) is always negative for x ≤ 2.

.. √(x² - 6x + 9) + √(2 - x) + (x - 3)
= √(x - 3)² + √(2 - x) + (x - 3)
= |x - 3| + √(2 - x) + (x - 3)
= -(x - 3) + √(2 - x) + (x - 3)
= √(2 - x)

The correct answer is A.
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