Question

Can someone check my work on radical expressions?

Answer 1

(√x + 1) + 5 = x
(√x + 1) + 5 - 5 = x - 5
(√x + 1) = x - 5
(√x + 1)^2 = (x - 5)^2
x + 1 = (x - 5)(x - 5)
x + 1 = x^2 - 5x - 5x + 25
x + 1 = x^2 - 10x + 25
x + 1 - x = x^2 - 10x + 25 - x
1 = x^2 - 11x + 25
1 - 1 = x^2 - 11x + 25 - 1
0 = x^2 - 11x + 24
(x - 3)(x - 8)
x - 3 = 0
x = 3
x - 8 = 0
x = 8
Solutions: 3, 8.
Checking:
(√3 + 1) + 5 = 3
(√4) + 5 = 3
2 + 5 = 3
7 = 3
This is a false statement.
(√8 + 1) + 5 = 8
(√9) + 5 = 8
3 + 5 = 8
8 = 8
This is a true statement.
3 is an extraneous solution.
8 is a working solution.

Answer 2

\[\sqrt{x+1}+5=x\]
is it your question
just want to make sure tthat one is under the square root ???

Answer 3

if 1 is under the square root then yes
your answer is right 3 doesn't work so thts mean 3 is extraneous solution
and 8 does work so your answer is right good job :)

Answer 4

your work looks correct.

the reason for the extra solution is that \(\sqrt{x+1}\) can give you a negative number as well. so if you take \(x=3\) then you would get:\[\sqrt{x+1}+5=x\]\[\therefore \sqrt{3+1}+5=3\]\[\therefore \sqrt{4}+5=3\]\[\therefore \pm2+5=3\]and now you can see that if we take the negative result of \(\sqrt{4}\) then it satisfies the equation.

however, in equations like this the squareroot usually implies the positive square root so we reject the solution \(x=3\)

Answer 5

you could also have solved this without having to square anything by first doing this:\[y=\sqrt{x+1}\]\[\therefore y+5=x\]\[\therefore y+6=x+1=y^2\]\[\therefore y^2-y-6=0\]\[\therefore (y-3)(y+2)=0\]\[\therefore y=3\text{ or }-2\]\[\therefore \sqrt{x+1}=3\text{ or }-2\]and at this point you could reject the \(-2\) solution as square roots are usually taken to be a positive value in these types of equations.

Answer 6

Thanks!

Answer 7

yw :)