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Math 222 Disc Wk1

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Submitted By donnacoreleon
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My first begins with so my expressions as given are x2-36 and 7w-2 3x 16w 2-
For my first expression x2-36 3x
I am trying to find the set of all real numbers that can fit as a solution without the denominator
Equaling to ‘0’ which is known as the domain any values that would cause x to equal ‘0’ are
Excluded values. If the value x were to be equal to ‘0’ it is no longer considered a ration expression and becomes irrational. For this first expression x cannot = 0 in set notation that would be
D= {x| xϵR, x≠0}

My second expression is 7w-2 16w2 -1

For this expression I would need to make the denominator equal to ‘0’ to find any excluded values for w

(4w+1)(4w-1)=0 Factors are set to equal 0

4w+1=0 and 4w-1= 0 add or subtract 1 from both sides

4w=-1 and 4w=1 divide each side by 4

W= -¼ or ¼ these are my excluded values

My solution set would be D={w| w є R, w≠ ± ¼ }

Both of my expressions have excluded values in the first one x cannot be equal to 0 because this would result in a denominator of ‘0’ which means I would no longer have a rational expression. In the second
Expression w cannot equal ± ¼ because this again would make the denominator0 making it into an irrational number and therefore undefined. My first expression only has one excluded value and my second expression contains
To values that are excluded.…...

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