How do you add #\sqrt { 16+ x } + \sqrt { 16- x } #?

1 Answer
George C.
Jan 27, 2018

#sqrt(16+x)+sqrt(16-x)# is already in simplest form.

Explanation:

Note first that it does make sense to multiply:

#sqrt(16+x) * sqrt(16-x) = sqrt((16+x)(16-x)) = sqrt(256-x^2)#

Addition is a very different case. There is no simpler form.

Under some circumstances it might be useful to "rationalise the numerator" as follows:

#sqrt(16+x)+sqrt(16-x) = ((sqrt(16+x)+sqrt(16-x))(sqrt(16+x)-sqrt(16-x)))/(sqrt(16+x)-sqrt(16-x))#

#color(white)(sqrt(16+x)+sqrt(16-x)) = ((sqrt(16+x))^2-(sqrt(16-x))^2)/(sqrt(16+x)-sqrt(16-x))#

#color(white)(sqrt(16+x)+sqrt(16-x)) = ((16+x)-(16-x))/(sqrt(16+x)-sqrt(16-x))#

#color(white)(sqrt(16+x)+sqrt(16-x)) = (2x)/(sqrt(16+x)-sqrt(16-x))#

...but this would be more useful the other way around, for example if you were asked to evaluate:

#lim_(x->0) (2x)/(sqrt(16+x)-sqrt(16-x)) = lim_(x->0) (sqrt(16+x)+sqrt(16-x)) = 4+4 = 8#

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