Answer:

c=6

Step-by-step explanation:

In order to solve the original equation, we would have to square both sides of the equation:

\begin{aligned}\sqrt{29+4w}&=23-cw\\\\ \left(\sqrt{29+4w}\right)^2&=(23-cw)^2\\\\ 29+4w&=(23-cw)^2\end{aligned}  

29+4w

​  

 

(  

29+4w

​  

)  

2

 

29+4w

​  

 

=23−cw

=(23−cw)  

2

 

=(23−cw)  

2

 

​  

 

However, squaring both sides of an equation can create extraneous solutions! [Why?]

Hint #22 / 4

Let’s plug \blueD w=\blueD{5}w=5start color #11accd, w, end color #11accd, equals, start color #11accd, 5, end color #11accd into the last equation we obtained:

\begin{aligned}29+4\blueD w&=(23-c\blueD{w})^2\\\\ 29+4(\blueD 5)&=(23-c(\blueD{5}))^2\\\\ 49&=(23-5c)^2\end{aligned}  

29+4w

29+4(5)

49

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=(23−cw)  

2

 

=(23−c(5))  

2

 

=(23−5c)  

2

 

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This equation is correct, both when 23-5c=723−5c=723, minus, 5, c, equals, 7 and when 23-5c=-723−5c=−723, minus, 5, c, equals, minus, 7.

However, the original equation is not correct for 23-5c=-723−5c=−723, minus, 5, c, equals, minus, 7, since this way we obtain \sqrt{49}=-7  

49

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=−7square root of, 49, end square root, equals, minus, 7.

Hint #33 / 4

Therefore, an extraneous solution is obtained for the ccc-value that makes 23-5c23−5c23, minus, 5, c equal -7−7minus, 7, which is c=6c=6c, equals, 6.

Substituting this back into the original equation gives \sqrt{29+4w}=23-6w  

29+4w

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=23−6wsquare root of, 29, plus, 4, w, end square root, equals, 23, minus, 6, w. You can now solve this for www and see for yourselves that w=5w=5w, equals, 5 is indeed extraneous.

Hint #44 / 4

The answer is:

c=6c=6

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