Could someone me with a few Algebra 2 problems?

[Timberizer]

I'm learning about imaginary numbers and radicals.

Here are the ones I need help with. Could someone explain how to do them, and what the answers are? Thanks

1. 2X + Yi = 3X + 1 + 3i (solve for X and Y)

2. (-4-5i)^2 (solve in a+bi form)

3. (1 - 2i)(-5 + 6i) (simplify in a +bi form)

4.
_____ _____
\/-32 + \/-50

5.

3___ 6_____
\/4 * \/3

6.

_______
\/ X + 7 = X-13 (solve for X if possible)

7.

5_________
/ X^10 Y^15
/_________
\/ 32

8.
________ ________
\/ 3X + 4 = \/ 2X + 3

 

[whozum]

1. The real parts are equal, the imaginary parts are equal.
2. Just foil it like you normally would, then when your done simplify your i's
3. Same thing as 2.
4. Write 32 and 50 out as products of their primes, ( 32 = 2*2*2*2*2), then take pairs of primes (2*2=4) and isolate them ([itex]\sqrt{32} = \sqrt{(2)(2)(2)(2)(2)} = \sqrt{(4)(2)(2)(2)} = 2\sqrt{(2)(2)(2)} [/itex]. Repeat until it is as simple as possible (you will be left with one term inside). From here, just use the definition of i to simplify the last bit. Do the same for 50.

[tex] 5. \frac{(3)(6)}{\sqrt{4}\sqrt{3}} [/tex] ? This ones easy, the denominator simplifies right away, no complex numbers involved. Dont forget to rationalize.

6. Square both sides
7. No idea what htis says
8. Square both sides

 

[Timberizer]

5.[tex] \frac{(3)(6)}{\sqrt{4}\sqrt{3}} [/tex] ? This ones easy, the denominator simplifies right away, no complex numbers involved. Dont forget to rationalize.

Thanks for the help, but you mixed up that problem. It was actually 3 root 4 times 6 root 3

 

[Timberizer]

#7 is actually: 5 root X^10 Y^15 over 32

 

[whozum]

5: 3rd root of 4? As in 4^(1/3) ? or 3 times square root of 4?

7: A square root can be expressed as ^(1/2), for example, square root of x^10 = (x^10)^(1/2). By properties of exponents this simplifies to x^5. Apply this to the rest.