# Math Help - Square Roots and Complex Numbers!!!

1. ## Square Roots and Complex Numbers!!!

Have a feq homework problems that i cant seem to solve
Multiply
1.(5 sq root over 2) (-4 sq root over 8)
2.(4 sq rt 2+ sq rt 3) (4 sq rt 2- sq rt 3)

Simplify
3.sq rt -12 x sq rt -9
4.(3-i) (6+3i)
5.(3-2i)/(4-sq rt-9)
6. (3+i) (4-2i)

Thanks for any help

2. Originally Posted by ls1ramair19
Have a feq homework problems that i cant seem to solve
Multiply
1.(5 sq root over 2) (-4 sq root over 8)
2.(4 sq rt 2+ sq rt 3) (4 sq rt 2- sq rt 3)

Simplify
3.sq rt -12 x sq rt -9
4.(3-i) (6+3i)
5.(3-2i)/(4-sq rt-9)
6. (3+i) (4-2i)

Thanks for any help
1.(5 sq root over 2) (-4 sq root over 8)
what are the first and second sqrts of? the 5 sq rt ? over 2 and the -4 sq root ? over 8

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

3.sq rt -12 x sq rt -9
sqrt(-12) * sqrt(-9)
= [sqrt(12)*sqrt(-1)] * sqrt(9)*sqrt(-1)
= sqrt(12) i * 3i
= 3sqrt(12) (i)^2 ......................................note, i^2 = -1
= -3sqrt(12)

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

(3-i) (6+3i) = 3(6) + 3(3i) - 6i - 3(i)^2
= 18 + 9i - 6i + 3
= 21 + 3i

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

6. (3+i) (4-2i)
same story as questions 2 and 4

(3+i) (4-2i) = 3(4) - 3(2i) + 4i - 2(i)^2
= 12 - 6i + 4i + 2
= 14 - 2i

3. Originally Posted by ls1ramair19
Have a feq homework problems that i cant seem to solve
Multiply
1.(5 sq root over 2) (-4 sq root over 8)
2.(4 sq rt 2+ sq rt 3) (4 sq rt 2- sq rt 3)
...
Hello,
Code:
5·√2·(- 4·√8) = 5·√2·(- 4·√(4*2))

= 5·√2·(- 4·2*√(2))= 5*(-4)*2*2 = -80
The second problem is a product of sum and difference:

(a + b)(a - b) = a² - b²
Code:
(4·√2 + √3)·(4·√2 - √3) = (4·√2)² - (√3)²

= 16*2 - 3 = 29
EB

4. Originally Posted by ls1ramair19
Multiply
1.(5 sq root over 2) (-4 sq root over 8)
o sorry, i didn't realize when you said "5 sq rt over 2" you meant the 5 * sqrt of 2.

saying "over" means divide, so even though you were refering to the fact the root sign goes "over" the 2, i got confused. anyway, earboth did it for you