Question

question
express in simplest radical form.
$sqrt{63}$

Explanation:

Step1: Factor the number inside radical

Find the prime - factorization of 63. $63 = 9\times7$.

Step2: Apply the square - root property

$\sqrt{63}=\sqrt{9\times7}$. According to the property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ ($a = 9$, $b = 7$ and $a\geq0$, $b\geq0$), we have $\sqrt{9\times7}=\sqrt{9}\cdot\sqrt{7}$.

Step3: Simplify the square - root of 9

Since $\sqrt{9}=3$, then $\sqrt{9}\cdot\sqrt{7}=3\sqrt{7}$.

Answer:

$3\sqrt{7}$