Question

How do you solve `(x + 1) ^ { 2} - 25= 0`?

Answer 1

`x=4, x=-6`

Explanation:

Put `y=x+1`, and thus `y^2-25=0`, which is zero when `y=+-5`, i.e. `(y+5)(y-5)=0`, and thus the roots as `y=+-5`

But `y=x+1` by definition; thus `x=4,` `x=-6`