User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

February 27th, 2018, 02:20 PM   #1
Newbie

Joined: Jan 2017
From: London

Posts: 10
Thanks: 0

Simplify simple expressions

Hi there,

Can someone please take a look at the attachment i.e. question 3.

I'm going through simplification processes and I understand the two principles to be:

1. Combine like terms
2. Factorize similar terms

I've applied this principle exercise 1 and 2 in the attachment but at quite a loss as to how to apply this principle to exercise 3.

A step by step procedure to the solution would be much appreciated.

Thanks
Attached Images Simplyfy.jpg (87.5 KB, 8 views) February 27th, 2018, 02:58 PM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 Rationalize the denominator. $\sqrt{a} + \dfrac{b}{\sqrt{a}} =$ $\sqrt{a} + \left ( 1 * \dfrac{b}{\sqrt{a}} \right )=$ $\sqrt{a}+ \left ( \dfrac{\sqrt{a}}{\sqrt{a}} * \dfrac{b}{\sqrt{a}} \right )=$ $\sqrt{a} + \dfrac{b\sqrt{a}}{\sqrt{a} * \sqrt{a}} =$ $\sqrt{a} + \dfrac{b\sqrt{a}}{a} =$ $\sqrt{a} * \left (1 + \dfrac{b}{a} \right ) = \sqrt{a} * \dfrac{a + b}{a}.$ Thanks from mathsonlooker February 27th, 2018, 06:24 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 Or $\displaystyle \sqrt{a} + \frac{b}{\sqrt{a}} = \frac{a}{\sqrt{a}} + \frac{b}{\sqrt{a}} = \frac{a + b}{\sqrt{a}}$, etc. Thanks from mathsonlooker March 5th, 2018, 04:19 AM   #4
Newbie

Joined: Jan 2017
From: London

Posts: 10
Thanks: 0

I appreciate the step by step follow through Quote:
 Originally Posted by JeffM1 Rationalize the denominator. $\sqrt{a} + \dfrac{b}{\sqrt{a}} =$ $\sqrt{a} + \left ( 1 * \dfrac{b}{\sqrt{a}} \right )=$ $\sqrt{a}+ \left ( \dfrac{\sqrt{a}}{\sqrt{a}} * \dfrac{b}{\sqrt{a}} \right )=$ $\sqrt{a} + \dfrac{b\sqrt{a}}{\sqrt{a} * \sqrt{a}} =$ $\sqrt{a} + \dfrac{b\sqrt{a}}{a} =$ $\sqrt{a} * \left (1 + \dfrac{b}{a} \right ) = \sqrt{a} * \dfrac{a + b}{a}.$ Tags expressions, simple, simplify Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post DecoratorFawn82 Abstract Algebra 3 September 1st, 2017 03:12 AM hkg277340 Pre-Calculus 4 December 3rd, 2016 12:08 PM Anon321 Algebra 2 July 26th, 2014 01:37 AM gregnolan Applied Math 2 October 13th, 2009 10:23 PM resch Algebra 1 September 29th, 2009 02:12 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      