what are the first and second sqrts of? the 5 sq rt ? over 2 and the -4 sq root ? over 81.(5 sq root over 2) (-4 sq root over 8)

[4sqrt(2) + sqrt(3)]*[4sqrt(2) - sqrt(3)]2.(4 sq rt 2+ sq rt 3) (4 sq rt 2- sq rt 3)

this is the foiled form of the difference of two squares, but if you don't see that, we can do it the hard way. we take one term in the first set of brackets and multiply each term in the second, then take the second term in the first pair of brackets and multiply each term in the second pair of brackets. here goes:

[4sqrt(2) + sqrt(3)]*[4sqrt(2) - sqrt(3)] = 4sqrt(2)*4sqrt(2) - 4sqrt(2)*sqrt(3) + sqrt(3)*4sqrt(2) - sqrt(3)*sqrt(3)

= [4sqrt(2)]^2 - 4sqrt(2)sqrt(3) + 4sqrt(2)sqrt(3) - [sqrt(3)]^2

= 16*2 - 3

= 29

sqrt(-12) * sqrt(-9)3.sq rt -12 x sq rt -9

= [sqrt(12)*sqrt(-1)] * sqrt(9)*sqrt(-1)

= sqrt(12) i * 3i

= 3sqrt(12) (i)^2 ......................................note, i^2 = -1

= -3sqrt(12)

following the same method as question 2, we get:4.(3-i) (6+3i)

(3-i) (6+3i) = 3(6) + 3(3i) - 6i - 3(i)^2

= 18 + 9i - 6i + 3

= 21 + 3i

(3 - 2i)/(4 - sqrt(-9)) = (3 - 2i)/(4 - sqrt(9)*sqrt(-1))5.(3-2i)/(4-sq rt-9)

= (3 - 2i)/(4 - 3i)

same story as questions 2 and 46. (3+i) (4-2i)

(3+i) (4-2i) = 3(4) - 3(2i) + 4i - 2(i)^2

= 12 - 6i + 4i + 2

= 14 - 2i