# Could someone me with some Algebra 2 problems?

• Timberizer
In summary, the conversation was about learning imaginary numbers and radicals. The person needed help with several problems involving solving for X and Y, simplifying expressions, and using the quadratic formula. They were given guidance on how to approach each problem and were offered further assistance if needed.
Timberizer

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

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

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

## What is Algebra 2?

Algebra 2 is a branch of mathematics that builds upon the concepts learned in Algebra 1. It covers topics such as polynomials, functions, systems of equations, and logarithms.

## Why do I need help with Algebra 2?

Algebra 2 can be a challenging subject for many students because it requires a strong understanding of fundamental algebraic concepts. Seeking help can improve your understanding and performance in the subject.

## What kind of problems can you help me with?

I can help you with a variety of Algebra 2 problems, including simplifying expressions, solving equations and inequalities, graphing functions, and working with exponential and logarithmic functions.

## How can you help me with my Algebra 2 problems?

I can provide step-by-step explanations and examples to help you understand the concepts and solve the problems. I can also offer tips and strategies for approaching and solving different types of problems.

## Can you guarantee that I will get the correct answers?

I cannot guarantee that you will get the correct answers, but I can guide you through the process and help you understand the concepts so that you can arrive at the correct answers on your own.

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