Question

Which value for the constant ccc makes w=5w=5w, equals, 5 an extraneous solution in the following equation? \sqrt{29+4w}=23-cw 29+4w ​ =23−cw

Answer 1

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

Answer 2

c = 6, makes w = 5 an extraneous solution of the equation √(29 + 4w) = 23 - cw.

What are extraneous solutions?

A root of a transformed equation that is not a root of the original equation because it was omitted from the domain of the original equation is referred to as an extraneous solution.

How to solve the question?

In the question, we are asked to find the value for the constant c, that makes w = 5, an extraneous solution in the equation:

√(29 + 4w) = 23 - cw.

We first try to solve this equation, by squaring both sides,

29 + 4w = (23 - cw)²,

or, 29 + 4w = 23² - 2(23)(cw) + (cw)²,

or, 29 + 4w = 529 - 46cw + c²w²,

Now, we substitute w = 5, in the above equation to get:

29 + 4(5) = 529 - 46c(5) + c²(5)²,

or, 29 + 20 = 529 - 230c + 25c²,

or, 25c² - 230c + 480 = 0.

Solving this by the quadratic formula, we get:

$c = \frac{-(-230)\pm \sqrt{(-230)^{2}-4(25)(480)}}{2(25)}$ ,

or, $c = \frac{230\pm \sqrt{52900-48000}}{50}$ ,

or, $c = \frac{230\pm \sqrt{4900}}{50}$ ,

or, $c = \frac{230\pm 70}{50}$ ,

or, c = 300/50 or 160/50,

or, c = 6 or c = 3.2.

Now, we will check these values in the original equation by keeping w = 5.

When c = 6:

√(29 + 4w) = 23 - cw,

or, √(29 + 4(5)) = 23 - (6)(5),

or, √(29 + 20) = 23 - 30,

or, √49 = -7,

or, 7 = -7, which is not true.

Hence, c = 6 makes w = 5 an extraneous solution.

When c = 3.2,

√(29 + 4w) = 23 - cw,

or, √(29 + 4(5)) = 23 - (3.2)(5),

or, √(29 + 20) = 23 - 16,

or, √49 = 7,

or, 7 = 7, which is true.

Hence, c = 3.2 makes w = 5 a proper solution.

Therefore, c = 6, makes w = 5 an extraneous solution of the equation √(29 + 4w) = 23 - cw.

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