Recent content by Astro

Assume you have a radial equation (eg. x+4 = √(x+10) ) that you want to solve for "x". To solve: \begin{align} (x+4 & = \sqrt[2] {x+10})^2 \nonumber \\ (x+4)^2 &= |x+10| \nonumber \end{align} For my question, we are only going to consider the case where x+10 < 0: \begin{align}...

Personal Question: Internet says the standardized math symbol for integers is ## \mathbb {Z}##. However, my Alberta MathPower 10 (Western Edition) textbook from 1998 says the symbol is I. I'm guessing that textbook is wrong? Or are both answers correct?

Appreciate the links. :smile: I'll read the articles and reply if I have any further questions.

*nods* That seems to make sense. I appreciate everyone's help. Thank you!

Well, the imaginary number in this case is extremely small, such that, that the answer is essentially -1 which is what it should be--so I now see what you're saying. But I'm confused as to why the calculator is displaying an imaginary component when the correct answer doesn't have one. I'm...

The answer I already posted is what it gave in float mode: ## -1 + 2\times 10^{-13}i ##. Or if you want exactly the same notation as the screen displayed it's -1+2E-13i.

Are you referencing the fact that: ##\sqrt[n] {x^n} = |x|## ? Because ##(\sqrt[n] {x})^n \neq \sqrt[n] {x^n}## when n is even and x < 0. But you probably already know that, so I'm not sure what you mean.

I'm not sure if I'm understanding your question but the calculator is on "float" mode (i.e. floating point).

Oh, you're right. I fixed the typo. Had typed it in wrong when using latex. It's the 2nd result that I was trying to type. I'm not concerned about the number of digits (accuracy) of the answer. Just not sure why the TI83+ is giving this answer in the first place.

I think the solution should be: METHOD #1: \begin{align} (\sqrt[4] {-1})^4 & = (-1)^{\frac 4 4} \nonumber \\ & = (-1)^1 \text{, can reduce 4/4 since base is a constant and not a variable in ℝ} \nonumber \\ & = -1 \nonumber \end{align} Alternatively, METHOD #2 for same answer is...

I like your example of solving ##x^2 = 4## by factoring to ##(x - 2)(x + 2) = 0## . However, how would you apply that approach to a question like ##(x+2)^{\frac 4 3}=16##? I don't know how to factor something like that. \begin{align} (\sqrt[3] {x+2})^4 & = 16 \nonumber \\ (x+2)^{\frac 4 3} & =...

Thank you for your most insightful reply. I enjoyed reading it. Really helped quite a bit. When you say that: Based on your reply, it seems there isn't a standard world-wide accepted interpretation of ##x ^{\frac m n}## for x ∈ ℝ . (Note: I do appreciate you explaining how most secondary...

Thank you for all your replies. It's helped a bit I still have some confusion. As mentioned, we know that for x ≥ 0 there is a power rule that states: ## x^{m/n} = \sqrt[n] {x^m} = (\sqrt[n] x)^m ## . Question #5a: For ## (\sqrt[n] x)^m##, it seems that you can only move m inside the...

(If I should have posted this in the Math thread instead of the Homework thread, please let me know.) I have three questions which I will ask in sequence. They all relate to each other. I've typed my questions and solutions attempts below. I've also attached a hand-written version of this...

I'm not sure I follow. To solve for the correct answer for "C" (which is C=4 according to the textbook), all I did was rearrange ## \frac{1}{2}Ln^2(y)=Ln|sec(x)|+C ## to solve for "y" and then subbing in the initial values x=0, and y=e^2 gives you ## e^2=e^{\sqrt{C}} ## from which you get C=4...