A question about absolute value signs and radicals, a specific problem.
- Thread starter Dale10101
- Start date
I agree with your work down to this line:
\(\displaystyle \displaystyle \sqrt{2y} × \left[ 8\ y\ |xy| +\ 42\ x\ y^2 \right] \)
But it is NOT necessarily true that \(\displaystyle y\ |y| = y^2 \)
It would appear that the workbook has assumed both x and y to be positive, so that the two terms can be added. Depending on the signs of x and y, the numeric coefficient could be ±50 or ±36. [And if y is negative, sqrt(y) is imaginary.] Unless there was something in the workbook to "guarantee" x any y both positive, then they are wrong and you are on a more rigorous track. (To be complete, you would need to show all four possible outcomes.)
Another question if I may.
Dr Phil, good medicine ...I see your point and Iwas wrong to blithely suck y into the absolute value braces, thanks.
Completed the problem as per the last statement that you identified as correct, slightly different results (?). Also attached a follow up question if I may. Thanks again, Dale
Dr Phil, good medicine ...I see your point and Iwas wrong to blithely suck y into the absolute value braces, thanks.
Completed the problem as per the last statement that you identified as correct, slightly different results (?). Also attached a follow up question if I may. Thanks again, Dale
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Your four cases should be identified by the signs of x and y. Did you remember that the sign of x is also relevant to the 2nd term>
\(\displaystyle 4 y \sqrt{8\ x^2\ y^3}\) is not a "statement," rather it is an "expression."
[An expression is just a noun, while a statement would have a verb such as "=".]
To say that this expression \(\displaystyle \in \Re \) does indeed require \(\displaystyle y \geq 0\). That is a piece of information that would not be available if you took the absolute value of y.