Question

how do you solve the square root of 2x minus the square root of x+1 equals 1?

Answer 1

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

Answer 2

yes that's the problem

Answer 3

you are going to have to square twice
i would start with
\[\sqrt{2x}=1+\sqrt{x+1}\] and then square both sides

Answer 4

\[(\sqrt{2x})^2=(1+\sqrt{x+1})^2\]\[2x=1+2\sqrt{x+1}+x+1\]

Answer 5

then would you combine like terms?

Answer 6

yes

Answer 7

that gives
\[2x=2+x+2\sqrt{x+1}\] then subtract \(x\) and \(2\) from both sides

Answer 8

you get
\[x-2=2\sqrt{x+1}\] and then you have to square again

Answer 9

i hate radicals lol

Answer 10

yeah it is a pain when there are two, because you have to square twice

Answer 11

next time you get
\[x^2-4x+4=4(x+1)\]

Answer 12

now you have a quadratic equation to solve
once you solve it, make sure to check your answers in the original equation, because they might not both work

Answer 13

would you take 4(x+1) to the other side

Answer 14

i would multiply out first, make it easier
\[x^2-4x+4=4x+4\]

Answer 15

then subtract from both sides, you get an easy equation to solve

Answer 16

how would you solve x squared -8x? would it be all that plus 1 or something?

Answer 17

factor and get
\[x^2=8x=0\]
\[x(x-8)=0\]

Answer 18

so \(x=0\) or \(x=8\)
don't forget to check them

Answer 19

o ok i get it now and thanks so much!!

Answer 20

yw