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Posted by T.Gracken on November 07, 2002 at 12:31:26:

In Reply to: Radicals posted by Mark on November 07, 2002 at 11:44:10:

: I have an algebra problem involving the simplification of radicals, which I have been unable to solve satisfactorily.

: The objective is to simplify and collect: SQRT[3x^2 - 18x + 27] - SQRT[27(x^2 + 2x + 1)].

: I checked by substituting 2 for x:
: SQRT[3(2)^2 - 18(2) + 27] - SQRT[27((2)^2 + 2(2) + 1)] = (-2(2) - 6)SQRT3

: I have been over this several times, and I am out of ideas. Where is my error?

: Thanks for the help.

This type of problem is an example of why we must be extremely careful in symbol manipulation.

It is also a good example to show why certain teachers should not assign them.

At first glance, it appears you did nothing wrong. (ask your teacher if he/she can find the error and see if they really know!!!)

The problem really comes down to the 'precise definition of a principal square root'. We tend to overlook an important part of the definition far too often.

That is: sqrt(A2) = |A|. {note that we usually disregard the absolute value and write sqrt(A2) = A which is not always correct}

so, for your procedure, i will use bold face font to indicate the necessary changes:

: My solution:

: SQRT[3x^2 - 18x + 27] - SQRT[27(x^2 + 2x + 1)]
: = SQRT[(3)(x^2 - 6x + 9)] - SQRT[(9)(3)(x^2 + 2x + 1)]
: = SQRT[(3)(x - 3)(x - 3)] - SQRT[(3^2)(3)(x + 1)(x + 1)]
: = SQRT[(3)(x - 3)^2] - SQRT[(3^2)(3)(x + 1)^2]
: = |x - 3|SQRT3 - |3(x + 1)|SQRT3
: = |x - 3|SQRT3 - |3x + 3|SQRT3
: = [|x - 3| - |3x + 3|]SQRT3

this can not be simplified further without splitting it into cases. that is, when is the absolute value positive and when is it negative.

If you have discussed implied domains then you might see that for the original problem, you must choose a value of x greater than or equal to 0 to stay in the realm of decimal numbers.

However, after simplifying, if you don't want to consider absolute values, then you must choose values of x greater than or equal to 3. If you had done this, everything would have seemed fine.

But, since you chose a value less than 3, a problem arises (unless you use the absolute value part of the definition).

notice that using the new result which includes the absolute values, you get the same answer regardless of which expression you substitute the value into. [that is, -8sqrt(3)]

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