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February 27th, 2018, 02:20 PM   #1
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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.

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February 27th, 2018, 02:58 PM   #2
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Rationalize the denominator.

$\sqrt{a} + \dfrac{b}{\sqrt{a}} =$

$\sqrt{a} + \left ( 1 * \dfrac{b}{\sqrt{a}} \right )=$

$\sqrt{a}+ \left ( \dfrac{\sqrt{a}}{\sqrt{a}} * \dfrac{b}{\sqrt{a}} \right )=$

$\sqrt{a} + \dfrac{b\sqrt{a}}{\sqrt{a} * \sqrt{a}} =$

$\sqrt{a} + \dfrac{b\sqrt{a}}{a} =$

$\sqrt{a} * \left (1 + \dfrac{b}{a} \right ) = \sqrt{a} * \dfrac{a + b}{a}.$
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February 27th, 2018, 06:24 PM   #3
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Or $\displaystyle \sqrt{a} + \frac{b}{\sqrt{a}} = \frac{a}{\sqrt{a}} + \frac{b}{\sqrt{a}} = \frac{a + b}{\sqrt{a}}$, etc.
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March 5th, 2018, 04:19 AM   #4
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I appreciate the step by step follow through
Quote:
Originally Posted by JeffM1 View Post
Rationalize the denominator.

$\sqrt{a} + \dfrac{b}{\sqrt{a}} =$

$\sqrt{a} + \left ( 1 * \dfrac{b}{\sqrt{a}} \right )=$

$\sqrt{a}+ \left ( \dfrac{\sqrt{a}}{\sqrt{a}} * \dfrac{b}{\sqrt{a}} \right )=$

$\sqrt{a} + \dfrac{b\sqrt{a}}{\sqrt{a} * \sqrt{a}} =$

$\sqrt{a} + \dfrac{b\sqrt{a}}{a} =$

$\sqrt{a} * \left (1 + \dfrac{b}{a} \right ) = \sqrt{a} * \dfrac{a + b}{a}.$
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