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
=(23−cw)
2
=(23−c(5))
2
=(23−5c)
2
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
=−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
=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
