Question

`m + 1 = √(4m + 49)`
​

Answer 1

First note that if
`m` is real-valued, then
`√(4m+49)` only exists if
`4m+49\ge0`, or
`m\ge-\frac{49}4=-12.25`.

Square both sides to get

`(m+1)^2=(√(4m+49))^2\implies m^2+2m+1=4m+49\implies m^2-2m-48=0`

This is easily factorized:

`m^2-2m-48=(m-8)(m+6)\implies m=8\text{ or }m=-6`

Both of these solutions are larger than
`-\frac{49}4`, so they are both valid solutions.