Question

How do you solve `11 + sqrt(2x+1)=2`?

It was suggested I post this

Answer 1

No solutions.

Explanation:

Our goal is to isolate `x` so we can find its value.

`11+sqrt(2x+1)=2`

`rArr sqrt(2x+1)=-9`

We see that the equation has no solutions because the radical operator over `RR` always returns positive values. So the equation has no solution.

Answer 2

The answer is no solution.

Explanation:

Your goal here is to isolate the variable.

First, start by subtracting 2 from both sides, as such:

`11-2+sqrt(2x+1)=2-2`

`9 +sqrt(2x+1) = 0`

Then subtract 9 from both sides:

`9-9 +sqrt(2x+1) = 0-9`

`sqrt(2x+1)=-9`

Now square both sides:

`(sqrt(2x+1))^2=(-9)^2`

`2x+1=81`

Then subtract 1 from both sides:

`2x+1-1=81-1`

`2x=80`

Finally, divided both sides by 2:

`(2x)/2=80/2`

`x=40`

Plug the answer into the original equation to check if it is correct:

`11+sqrt(2(40)+1)=2`

`11+9=2`

`20!=2`

When plugged in, we can see that `x=40` is not correct, so this problem has no solution.

Answer 3

No Real Solution

Explanation:

Begin by subtracting `11` from both sides

`cancel(11color(red)(-11))+sqrt(2x+1)=2color(red)(-11)`

`sqrt(2x+1)=-9`

Square both sides to eliminate the radical on the left side

`sqrt(2x+1)^color(red)2=-9^color(red)2`

`2x+1=81`

Subtract `1` from both sides

`2x+cancel(1color(red)(-1))=81color(red)(-1)`

`2x=80`

Divide both side by `2`

`cancel2/cancelcolor(red)2x=80/color(red)2`

`x=40`

However if we verify this by substituting `40` for `x` in the original expression we'll find:

`11+sqrt(2(color(red)40)+1)=2`

`11+sqrt(81)=2`

`11+9=2`

`20!=2`

`:.` There are no real solutions

Answer 4

No solution for principle root `->+sqrt(2x+1)+11=2`

The convention is; that unless the question states
`+-sqrt("something")` we only use the `+sqrt("smething")` which is the principle root.

Explanation:

However, just for fun; lets investigate the scenario of `(-2)xx(-2)=+4=(+2)xx(+2)`

So `sqrt(4)=+-2" is definitely viable "ul("only if you ignore the convention")`

The convention states :`sqrt(4)" can only "=+2" but you need"+-sqrt(4)" to have "+-2`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given: `sqrt(2x+1)+11=2`

Write as `+-sqrt(2x+1)+11=2 color(red)(larr" not the actual question")`

Subtract 11 from both sides

`+-sqrt(2x+1)=-9`

Clearly `+sqrt(2x+1)=-9` does not have a solution for `x in RR` as each side of the equals is the opposite sign.

However `-sqrt(2x+1)=-9` may have a solution as both sides are negative. Accepting this should not be done as it is against the convention.`color(red)(" Note that this is not the question as given")`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Just for fun consider "-sqrt(2x+1)=-9)`

Square both sides

`2x+1 = +81`

`2x=80`

`x=40`
Which is not the the answer for the given question as written.

The graph clearly shows no solution for `+sqrt(2x+1)=-9` which is the answer to the given question as written.

If the question had been `+-sqrt(2x+1)` then we could legitimately have stated the answer as being: `(x,y)->(40,9)` is a solution for `-sqrt(2x+1)=-9`