# Could someone please help me with some Algebra 2 problems?

1. May 12, 2005

### 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

2. May 13, 2005

### PhysicsinCalifornia

1)$$2x + yi = 3x + 1 + 3i$$
A little nasty, but try to solve for y first. Then try solving for x

2)$$(-4-5i)^2 = (-4-5i)(-4-5i)$$
It's simple algebra from there(and remember, $$i^2$$ = -1)

3)$$(1 - 2i)(-5 + 6i)$$
Same as #2

4)$$\sqrt{-32} + \sqrt{-50}$$
I'm assuming you tried to be fancy and make the problem look visually like the one here.
I think your book should have a rule on if there is a negative in the square root. $$\sqrt{-x} = i\sqrt{x}$$. Use this rule to solve

5)$$\sqrt[3]{4} \sqrt[6]{3}$$ (This is what you mean?)
Can you use a calculator? or you have to look up the rule to solve this?

6)$$\sqrt{x + 7} = x - 13$$
Try to get rid of that squareroot first. Careful, make sure you do it for both sides!

7)$$\frac{\sqrt[5]{x^{10} y^{15}}}{\sqrt{32}}$$ Is this what you mean?

8)$$\sqrt{3x + 4} = \sqrt{2x + 3}$$
Same as #6

3. May 16, 2005

### z-component

I will solve number 6 for you so you can see if you're on the right track.

The strategy is finding values for x. The first thing you need to do is get rid of the square root sign. To do this, we need to square both sides. Squaring the side with the square root will cancel out the sign and leave what's underneath it. Squaring the binomial on the other side requires "FOILing," or simply multiplying them to make a quadratic polynomial. Next, you need to make the quadratic polynomial equal to zero so you can use the quadratic formula or factor to find values for x. When you are left with the quadratic polynomial equal to zero, you may simply find x.

$$\sqrt{x+7}=(x-13)$$

$$(\sqrt{x+7})^2=(x-13)^2$$

= $$x+7=(x-13)(x-13)$$

= $$x+7=x^2-26x+169$$

= $$x^2-27x+162=0$$

Finally, we find the real solutions are

$$x=\{18,9\}$$

Follow that general strategy for number 8 as well. Let us know if you need more assistance. We'll be glad to help you with any others.

z-c

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