Question 1207483
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Yes you can use the FOIL method to expand out the radicand.
The distributive rule or box method are two other alternative pathways.
I'll use the distributive rule. 
(4+3i)(3i-4)
= 4(3i-4)+3i(3i-4)
= 12i-16+9i^2-12i
= -16+9i^2
= -16+9(-1)
= -25


In short,
(4+3i)(3i-4) = -25


Then,
sqrt( (4+3i)(3i-4) )
= sqrt(-25)
= sqrt(-1*25)
= sqrt(-1)*sqrt(25)
= i*5
= <font color=red>5i</font>



Therefore,
sqrt( (4+3i)(3i-4) ) = <font color=red>5i</font>


Confirmation using WolframAlpha
<a href="https://www.wolframalpha.com/input?i=sqrt%28+%284%2B3i%29%283i-4%29+%29+%3D+5i">https://www.wolframalpha.com/input?i=sqrt%28+%284%2B3i%29%283i-4%29+%29+%3D+5i</a>


Edit:
{{{sqrt(-25) = ""+- 5i}}} is NOT correct
Instead it should be {{{sqrt(-25) = 5i}}}
I've seen students make this mistake many times. 
The plus/minus is not inherently built into the square root function. 
The plus/minus only results when solving quadratics.
There's a difference between solving something like {{{x^2-4 = 0}}} which has two solutions vs something like {{{x = sqrt(4) = 2}}} which has one solution.
The square root function produces ONE output for any given input.
It's a bit shocking that tutor ikleyn has made such an error.
WolframAlpha is correct.
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