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

Jan 22, 2016

$r = 7$

#### Explanation:

This is a perfect square trinomial—it is equal to

${\left(r - 7\right)}^{2} = 0$

This can be verified by redistributing ${\left(r - 7\right)}^{2}$:

$\left(r - 7\right) \left(r - 7\right) = {r}^{2} - 7 r - 7 r + 49 = {r}^{2} - 14 r + 49$

Back to solving the equation.

${\left(r - 7\right)}^{2} = 0$

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 $7$.