Question

X+2 = √3x + 10
What is the solution and the extraneous solution. Please show me how you got the answer so I can understand.
Also, Please explain what an extraneous solution is
Thanks

Answer 1

First, an extraneous solution is a solution which doesn't work :-)

When you take the square root, you actually have two different values. For example, the square root of 4:

\[\sqrt{4} = 2\]because \(2*2=4\)
but that isn't the only value that makes a true statement:
\[\sqrt{4} = -2\]because \(-2*-2 = 4\) so really \[\sqrt{4} = \pm 2\]and by convention we take the positive value.

Answer 2

Extraneous solutions come about when we have to solve an equation having a radical sign (\(\sqrt{}\)) in it. We typically square the equation while solving, and the "other" value of the square root contributes to making the extraneous solution, but when you plug it into the original formula, you get a false statement, such as \(3=4\).

Answer 3

To solve a problem like this one, isolate the term with the radical on one side of the equation.

\[x+2 = \sqrt{3x} + 10\]\[x+2 - 10 = \sqrt{3x}\]\[x-8 = \sqrt{3x}\]Now square both sides and solve for \(x\). Try out both values you get, one will work (the solution), and one will not (the extraneous solution).

Answer 4

Just to clarify, the problem is

\(x+2 = \sqrt{3x + 10} \)

not

\(x+2 = \sqrt{3x} + 10 \)

Correct?

Answer 5

@mathstudent55 good point, your reading probably is more likely!

Answer 6

@whpalmer4 Your explanation is excellent, as usual, and I'd hate to see you waste time solving the wrong equation because, also as usual, the post is not clear.

Answer 7

Having solved it both ways, your reading has a simpler solution, so probably is the right one.

Thanks for the kind words!

Answer 8

My pleasure.

Answer 9

Yes it is supposed to be \[x+2=\sqrt{3x+10}\]

Answer 10

Thanks for the update, @royalbug. Shall we work the problem, or check your answer?

\[x+2 = \sqrt{3x+10}\]We have the radical isolated already, so we just square both sides of the equation:
\[(x+2)^2 = (\sqrt{3x+10})^2\]Expanding that,
\[(x+2)(x+2) = 3x+10\]\[x^2+4x+4 = 3x+10\]
Simplify and solve for \(x\)...

Answer 11

I got
x = 2, -3

Answer 12

I agree with your work so far. Now try plugging those solutions into the original equation. Do they both satisfy the original equation?

Answer 13

2 does but -3 does not

Answer 14

Very good. Which one is the extraneous solution?

Answer 15

-3?

Answer 16

As an old math teacher of mine used to say when someone would answer with a questioning tone in his voice, "is that an answer, or a prayer?" :-)

Yes, -3 is the extraneous solution, because \[(-3) + 2 = \sqrt{3(-3)+10}\]\[-1 = \sqrt{1}\]\[-1=1\]is not true.

Any questions about this?

Answer 17

Nope. Thank you so much. I appreciate your help!

Answer 18

Excellent! This is a fine example of why you should always check all of your answers when working a problem that generates multiple solutions.

Answer 19

Now, I have to create my own extraneous equation using the formula \[x+a = \sqrt{bx + c}\]. Where a, b, and c are all positive integers and b >1. The solution
has to equal 7 and there must be an extraneous solution. Do you know how I would do that?

Answer 20

Hi @royalbug, do you have any ideas about how to tackle this?

Answer 21

well, I know i need to find integers for a, b, and c but other than that not really

Answer 22

Sorry, been having some internet problems here. Are you still around? @royalbug

Answer 23

I would work the second part of this problem by plugging in the solution in the prototype equation.

\[x+a = \sqrt{bx+c}\]\[x=7\]\[7+a = \sqrt{b(7)+c}\]Now pick values of \(b\) and \(c\) that make the radical expression easy to evaluate, and satisfy the conditions that \(b>1\) and \(b,c\) are both positive integers. From there, solve for \(a\) and you're done! Solve the equation to find the extraneous solution (and make sure the "real" solution" turns out to be 7, of course!)