How do you evaluate #\sqrt { - 81} + \sqrt { 49}#?

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Answer 1 · 2017-03-17T06:41:23

#sqrt(-81)+sqrt49=9i+7#

Here we are dealing with square roots of negative numbers i.e. imaginary and / or complex numbers are involved. Hence recall that #i^2=-1#

Therefore #sqrt(-81)+sqrt49#

= #sqrt(-1xx9xx9)+sqrt(7xx7)#

= #sqrt(i^2xx9^2)+sqrt(7^2)#

= #9i+7#

Answer 2 · 2017-03-17T20:13:00

#sqrt(-81)+sqrt(49) = 9i+7#

A square root #a# of a number #n# is a number such that #a^2=n#.

Every non-zero number #n# has two square roots #sqrt(n)# and #-sqrt(n)#.

The #sqrt# symbol denotes the principal square root which is defined as follows:

If #n > 0# then #sqrt(n)# is the positive square root.

If #n < 0# then #sqrt(n) = i sqrt(-n)# where #i# is the imaginary unit.

In our example:

#sqrt(-81)+sqrt(49) = sqrt(-9^2)+sqrt(7^2)#

#color(white)(sqrt(-81)+sqrt(49)) = i sqrt(9^2)+sqrt(7^2)#

#color(white)(sqrt(-81)+sqrt(49)) = 9i+7#

#color(white)()#
Square roots of complex numbers

Every non-zero complex number #z# can be uniquely represented in the form:

#z = r(cos theta + i sin theta)#

for some #r > 0# and #-pi < theta <= pi#

Then the principal square root is:

#sqrt(z) = sqrt(r)(cos (theta/2) + i sin (theta/2))#

This is consistent with the definitions of square roots of real numbers described above.