My Math Forum Math Homework Help "Extend and Challenge" Questions
 User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 September 11th, 2013, 04:27 PM #1 Member   Joined: Sep 2013 Posts: 58 Thanks: 0 Math Homework Help "Extend and Challenge" Questions Hey, I'd really appreciate help on these questions and how to solve them. For the first one, I had no idea how to do it (Q#13). On the second one, this is what I got up too (Q#14). 1. Cube each Side - m+9 = 27 + 27THIRDROOTOF(m-9) + 9(2/3)ROOTOF(m-9) + m-9 2. Rearrange and Sub x for THIRDROOTOF(m-9) - 0=9+27x+9x^2 3. Divide both sides by 9 0 = 1 + 3x + x^2 4. (Here I got lost so I just tried to do stuff to help me out.) 0 = 1 + 3x +x^2 -x +x 0 = 1 + 2x + x^2 + x 0 = (1+x)^2 + x http://img194.imageshack.us/img194/1097/kvgc.jpg
 September 11th, 2013, 05:32 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,842 Thanks: 1068 Math Focus: Elementary mathematics and beyond Re: Math Homework Help "Extend and Challenge" Questions $(m\,+\,9)^{\frac13}\,=\,3\,+\,(m\,-\,9)^{\frac13} \\ (m\,+\,9)^{\frac13}\,-\,(m\,-\,9)^{\frac13}\,=\,3 \\ m\,+\,9\,-\,3(m\,+\,9)^{\frac23}(m\,-\,9)^{\frac13}\,+\,3(m\,+\,9)^{\frac13}(m\,-\,9)^{\frac23}\,-\,m\,+\,9\,=\,27 \\ 3(m\,+\,9)^{\frac13}(m\,-\,9)^{\frac13}$(m\,-\,9)^{\frac13}\,-\,(m\,+\,9)^{\frac13}$\,=\,9 \\ -9(m\,+\,9)^{\frac13}(m\,-\,9)^{\frac13}\,=\,9 \\ (m\,+\,9)^{\frac13}(m\,-\,9)^{\frac13}\,=\,-1\\ (m\,+\,9)(m\,-\,9)\,=\,-1 m^2\,-\,81\,=\,-1\,\Rightarrow\,|m|\,=\,4\sqrt{5}$
September 11th, 2013, 07:38 PM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 407

Re: Math Homework Help "Extend and Challenge" Questions

Hello, Tzad!

Quote:
 $\text{13. Determine the exact value: }\:\sqrt{10\,+\,\sqrt{10\,+\,\sqrt{10\.+\.\text{. . .}}}}$

$\text{W\!e have: }\;x \;=\;\sqrt{10\,+\,\sqrt{10\,+\,10\,+\,\text{. . .}}}$

[color=beige]. . . . . . . . [/color]$x \;=\;\sqrt{10\,+\,\underbrace{\sqrt{10\.+\.\sqrt{1 0\,+\,\text{. . .}}}}_{\text{This is }x}}$

[color=beige]. . . . . . . . [/color]$x \;=\;\sqrt{10\,+\,x}$

$\text{Square: }\;x^2 \;=\;10\,+\,x \;\;\;\Rightarrow\;\;\; x^2 \,-\,x\,-\,10 \:=\:0$

$\text{Quadratic Formula: }\;x \;=\;\frac{-(-1)\,\pm\,\sqrt{(-1)^2\,-\,4(1)(-10)}}{2(1)} \;=\;\frac{1\,\pm\,\sqrt{41}}{2}$

$\text{Since }x\text{ is positive: }\:x \;=\;\frac{1\,+\,\sqrt{41}}{2}$

September 13th, 2013, 02:14 PM   #4
Member

Joined: Sep 2013

Posts: 58
Thanks: 0

Re: Math Homework Help "Extend and Challenge" Questions

Quote:
 Originally Posted by greg1313 $(m\,+\,9)^{\frac13}\,=\,3\,+\,(m\,-\,9)^{\frac13} \\ (m\,+\,9)^{\frac13}\,-\,(m\,-\,9)^{\frac13}\,=\,3 \\ m\,+\,9\,-\,3(m\,+\,9)^{\frac23}(m\,-\,9)^{\frac13}\,+\,3(m\,+\,9)^{\frac13}(m\,-\,9)^{\frac23}\,-\,m\,+\,9\,=\,27 \\ 3(m\,+\,9)^{\frac13}(m\,-\,9)^{\frac13}$(m\,-\,9)^{\frac13}\,-\,(m\,+\,9)^{\frac13}$\,=\,9 \\ -9(m\,+\,9)^{\frac13}(m\,-\,9)^{\frac13}\,=\,9 \\ (m\,+\,9)^{\frac13}(m\,-\,9)^{\frac13}\,=\,-1\\ (m\,+\,9)(m\,-\,9)\,=\,-1 m^2\,-\,81\,=\,-1\,\Rightarrow\,|m|\,=\,4\sqrt{5}$
Hey Greg! Thanks for the help!

I wanted to know what you did between steps 4 and 5.

Thanks,
Tzad

 September 13th, 2013, 02:17 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,842 Thanks: 1068 Math Focus: Elementary mathematics and beyond Re: Math Homework Help "Extend and Challenge" Questions $\text{If }(m\,+\,9)^{\frac13}\,=\,3\,+\,(m\,-\,9)^{\frac13}\text{ then }{(m\,-\,9)^{\frac13}\,-\,(m\,+\,9)^{\frac13}\,=\,-3$ Does that help?
September 13th, 2013, 05:59 PM   #6
Global Moderator

Joined: Dec 2006

Posts: 19,293
Thanks: 1684

Quote:
 Originally Posted by Tzad 0 = 1 + 3x + x^2
0 = 1 + x(3 + x)
x³(3 + x)³ = -1
(m - 9)(m + 9) = -1
m² = 81 - 1 = 80
|m| = 4?5

 September 14th, 2013, 10:32 AM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,912 Thanks: 883 Re: Math Homework Help "Extend and Challenge" Questions $(m\,+\,9)^{\frac13}\,-\,(m\,-\,9)^{\frac13}\,=\,3 \\$ To Greg and Skip....I'm confused and hope to learn something here... I followed your work to arrive at m = 4SQRT(5), but that I'm quite sure is not the solution for the above equation... http://www.wolframalpha.com/input/?i=%2 ... +-+3+%3D+0 So what did I miss? Thanks guys!
September 14th, 2013, 12:43 PM   #8
Member

Joined: Sep 2013

Posts: 58
Thanks: 0

Re: Math Homework Help "Extend and Challenge" Questions

Quote:
 Originally Posted by greg1313 $\text{If }(m\,+\,9)^{\frac13}\,=\,3\,+\,(m\,-\,9)^{\frac13}\text{ then }{(m\,-\,9)^{\frac13}\,-\,(m\,+\,9)^{\frac13}\,=\,-3$ Does that help?
Thank you!

It helped me a lot.

September 14th, 2013, 12:51 PM   #9
Global Moderator

Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,842
Thanks: 1068

Math Focus: Elementary mathematics and beyond
Re: Math Homework Help "Extend and Challenge" Questions

Quote:
 Originally Posted by Denis So what did I miss?
I don't think that would be the first time W|A has been wrong. Check it with a calculator.

September 14th, 2013, 12:59 PM   #10
Member

Joined: Sep 2013

Posts: 58
Thanks: 0

Re: Math Homework Help "Extend and Challenge" Questions

Quote:
Originally Posted by greg1313
Quote:
 Originally Posted by Denis So what did I miss?
I don't think that would be the first time W|A has been wrong. Check it with a calculator.
I want to wire you money for helping me.

What is your paypal email?

Search tags for this page

### challenge and extend algebra

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post ryanninjasheep Applied Math 2 January 19th, 2014 07:55 PM SedaKhold Calculus 0 February 13th, 2012 11:45 AM The Chaz Calculus 1 August 5th, 2011 09:03 PM Aggie10 Algebra 12 September 11th, 2010 07:27 PM katie0127 Advanced Statistics 0 December 3rd, 2008 01:54 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2018 My Math Forum. All rights reserved.