Question

Question #07e50

Answer 1

`x=3`. Don't be fooled by a phantom(see below).

Explanation:

You have to square the formula to remove `sqrt` sign, but if
you calculate `(2x+sqrt(x+1))^2` will yield `4x^2+4xsqrt(x+1)+x+1` and make matters worse.

First, move the term `2x` to the right side:
`sqrt(x+1)=8-2x`

Then square the formula:
`(sqrt(x+1))^2=(8-2x)^2`
`x+1=64-32x+4x^2`

And reduce this:
`4x^2-33x+63=0`

Use the quadratic formula and you will reach
`x=(33±sqrt(33^2-4xx4xx63))/8`
`x=(33±9)/8`
`x=21/4,3`
(or you can factor the equation to `(4x-21)(x-3) =0`)
Important: This is not the final answer and you need to check it!

As the range of `sqrt(x+1)` is `sqrt(x+1)>= 0` , the condition
`8-2x> 0` must be satisfied. `x=3` meets the
condition but `x=21/4 = 5.25` does not.

Therefore the root of this equation is `x=3`.