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a) What is the solution to the equation?

 

 

b) What is the extraneous solution? Why?

 

 

c) In general, what is an extraneous solution?

 

 

 

(please show all work, this genuinely helps me understand how you got to the answer, thank you in advance!)

 Apr 24, 2021
 #1
avatar+509 
+5

Equation is   \(x+2={\sqrt{3x+10}}\)

 

a) First of all, squaring both sides

\((x+2)^2=3x+10\)

\(x^2+4x+4=3x+10\)

\(x^2+x-6=0\)

⇒ \(x^2+3x-2x-6=0\)

\((x+3)(x-2)=0\)

 

∴ x = -3 or 2

 

 

b) Here we can't have \({\sqrt{3x+10}}<0\)

∵ Any value  \(x<{-10 \over 3}\)  is extraneous. 

 

∴ Squaring both sides in  \({\sqrt {3x+10}}=x\)

\(3x+10=x^2\)

\(x^2-3x-10=0\)

⇒ \((x-5)(x+2)=0 \)

\(x=5\) or \(-2\)

 

1st case : Putting x=5 in eq.                          2nd case: Putting x=-2 in eq. 

\({\sqrt{3(5)+10}} = 5\)                                       \({\sqrt{3(-2)+10}}=(-2)^2\)

⇒ \(5=5\)                                                     ⇒ \(2≠-2\)

 

∴ x = 5 is the extraneous solution. 

 

 

c) An extraneous solution is a solution that's evolved from the process of solving of the equation but its not a valid solution to the equation. You know, they are those "extra answers" that we get while solving the problem and they may not necessarily be valid

 

 

~Roll the credits 

 Apr 24, 2021
 #2
avatar+79 
+2

This is perfect, thank you so much for the detailed explanation! I appreciate it :)

Corduroy  Apr 24, 2021
 #3
avatar+509 
+2

You're very welcome! :D 

amygdaleon305  Apr 24, 2021

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