Question

Solve for x, given the equation Square root of x minus 5 + 7 = 11.
x = 21, solution is extraneous x = 21, solution is not extraneous x = 81, solution is extraneous x = 81, solution is not extraneous

Answer 1

\[\sqrt{x-5}+7=11\]

Answer 2

We can first subtract on both sides so that we have isolated the squareroot

Answer 3

that might've been what I did wrong I started off doing 11^2
\[\sqrt{x-5}=4\]

Answer 4

Now if you square both sides, you will result with:

\(\sf (\sqrt{x-5})^2 = 4^2\\x-5= 16\)

Answer 5

then add 5 to both sides thatll be x=21 now whats an extraneous solution

Answer 6

An extraneous solution is one that will not work if you plug it back in.

So we need to test if x = 21 actually works.

Answer 7

Recall that you will obtain TWO possible solutions; it's up to you to determine which one is a solution and which is an extraneous solution.

Answer 8

\[\sqrt{21-5}=4\]

Answer 9

That means it is not an extraneous solution. :)

Answer 10

wouldn't it be extraneous? 21-5=16 then square root is 4

Answer 11

and then 4 + 7 = 11

Answer 12

http://hotmath.com/hotmath_help/topics/extraneous-solutions.html

"An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation."

Answer 13

So it's not?

Answer 14

Yes, it is not extraneous.

Answer 15

okay thank you for your help

Answer 16

Anytime! :)