How do you simplify #sqrt(-15)( sqrt(-4 -sqrt7))#?

1 Answer
George C.
Apr 16, 2017

#sqrt(-15)sqrt(-4-sqrt(7)) = -sqrt(30)/2-sqrt(210)/2#

Explanation:

By definition and convention, the principal square root of a negative real number #x# is given by:

#sqrt(x) = isqrt(-x)#

where #i# is the imaginary unit, satisfying #i^2=-1#.

If #a, b >= 0# then:

#sqrt(ab) = sqrt(a)sqrt(b)#

As a first step in attempting to simplify the given expression, let's try to find a simpler expression for:

#sqrt(4+sqrt(7))#

This may be expressible in the form #a+bsqrt(7)#. Let's see what we get when we square #a+bsqrt(7)#...

#(a+bsqrt(7))^2 = (a^2+7b^2)+2ab sqrt(7)#

So we would like #a^2+7b^2 = 4# and #2ab=1#

Notice that if #a=b=1# then we would get #a^2+7b^2=8# and #2ab=2#, both twice the wanted values.

So we can use #a=b=sqrt(2)/2# to find:

#sqrt(4+sqrt(7)) = sqrt(2)/2 + (sqrt(2)sqrt(7))/2 = sqrt(2)/2+sqrt(14)/2#

From our previous remarks:

#sqrt(-4-sqrt(7)) = isqrt(4+sqrt(7)) = i(sqrt(2)/2+sqrt(14)/2)#

Then:

#sqrt(-15)sqrt(-4-sqrt(7)) = isqrt(15)*i(sqrt(2)/2+sqrt(14)/2)#

#color(white)(sqrt(-15)sqrt(-4-sqrt(7))) = -sqrt(15)(sqrt(2)/2+sqrt(14)/2)#

#color(white)(sqrt(-15)sqrt(-4-sqrt(7))) = -sqrt(30)/2-sqrt(210)/2#

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