Question

How do you find all the real and complex roots of `r²-14r+49=0`?

Answer 1

`r=7`

Explanation:

This is a perfect square trinomial—it is equal to

`(r-7)^2=0`

This can be verified by redistributing `(r-7)^2`:

`(r-7)(r-7)=r^2-7r-7r+49=r^2-14r+49`

Back to solving the equation.

`(r-7)^2=0`

Now, we can take the square root of both sides.

`r-7=0`

`r=7`

This is the only answer. Even though quadratics typically have two answers, this has a root of `7` with a multiplicity of `2`, which means that both its roots are `7`.