Question

Can somebody please help and explain to me how to check if the solution is extraneous?

Answer 1

\[\sqrt{2x+6}-\sqrt{x-1}=2\]

Answer 2

i would add \(\sqrt{x-1}\) to both sides and start with
\[\sqrt{2x+6}=\sqrt{x-1}+2\] then square both sides

Answer 3

then, although your teacher probably wants you to do something else, i would guess at a solution, which should be easy because there are not too many ways to pick an \(x\) make both \(x-1\) and also \(2x+6\) to be perfect squares

Answer 4

try \(5\) and see that \(2\times 5+6=16\) is a perfect square (it is the square of \(4\)) and also \(5-1=4\) is also a perfect square
got it on the first try

Answer 5

by checking your answer
"extraneous" is another name for a solution that doesn't actually work

Answer 6

a solution that doesn't work with the original equation

Answer 7

for example \(5\) is a solution to
\[\sqrt{2x+6}-\sqrt{x-1}=2\] because
\[\sqrt{2\times 5+6}-\sqrt{5-1}=2\\
\sqrt{16}-\sqrt{4}=2\\
4-2=2\] is true

Answer 8

Okay thank you sooo much!!!

Answer 9

yw

Answer 10

I am public schooled!

Answer 11

oh ok lol