Question

simplify.
$6\sqrt{63}$

Explanation:

To simplify \(\sqrt{63}\), we can factor the radicand (the number inside the square root) into a product of a perfect square and another number.

Step 1: Factor the radicand

First, find the prime factorization of \(63\):
\(63 = 9 \times 7\)

Notice that \(9\) is a perfect square (\(9 = 3^2\)).

Step 2: Apply the square root property

The square root of a product is the product of the square roots:
\(\sqrt{ab} = \sqrt{a} \times \sqrt{b}\) (for \(a \geq 0, b \geq 0\))

Applying this to \(\sqrt{63}\):

$$ \sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7} $$

Step 3: Simplify the perfect square root

Since \(\sqrt{9} = 3\) (because \(3^2 = 9\)):

$$ \sqrt{63} = 3\sqrt{7} $$

So the simplified form of \(\sqrt{63}\) is \(3\sqrt{7}\).

Answer:

To simplify \(\sqrt{63}\), we can factor the radicand (the number inside the square root) into a product of a perfect square and another number.

Step 1: Factor the radicand

First, find the prime factorization of \(63\):
\(63 = 9 \times 7\)

Notice that \(9\) is a perfect square (\(9 = 3^2\)).

Step 2: Apply the square root property

The square root of a product is the product of the square roots:
\(\sqrt{ab} = \sqrt{a} \times \sqrt{b}\) (for \(a \geq 0, b \geq 0\))

Applying this to \(\sqrt{63}\):

$$ \sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7} $$

Step 3: Simplify the perfect square root

Since \(\sqrt{9} = 3\) (because \(3^2 = 9\)):

$$ \sqrt{63} = 3\sqrt{7} $$

So the simplified form of \(\sqrt{63}\) is \(3\sqrt{7}\).