Question

What are the beginning steps of solving this?

`2x + √(x + 1) = 8`

Answer 1

Isolate the radical term, then square both sides:

`2x+√(x+1)=8`

`√(x+1)=8-2x`

Note that
`√(x+1)` is non-negative and requires that
`x\ge-1` in order to be defined, while
`8-2x<0` when
`x>4`. This means we can only find real solutions in the range
`-1\le x\le4`.

`(√(x+1))^2=(8-2x)^2`

`x+1=(8-2x)^2`

Now just expand the RHS and simplify as much as possible:

`x+1=64-32x+4x^2`

`4x^2-33x+63=0`

`(x-3)(4x-21)=0`

`\implies x=3,x=\frac{21}4`

Now,
`\frac{21}4=5.25`, which is larger than 4, so we omit that solution. Then
`x=3` is the only solution.