Question

radical expressions
simplify
\sqrt{63x}

Explanation:

Step1: Factor the radicand

Factor 63 into prime factors: \( 63 = 9\times7 \), so \( \sqrt{63x}=\sqrt{9\times7\times x} \)

Step2: Apply the square - root property

Using the property \( \sqrt{ab}=\sqrt{a}\cdot\sqrt{b} \) (\( a\geq0,b\geq0 \)), we have \( \sqrt{9\times7\times x}=\sqrt{9}\cdot\sqrt{7x} \)
Since \( \sqrt{9} = 3 \), then \( \sqrt{9}\cdot\sqrt{7x}=3\sqrt{7x} \)

Answer:

\( 3\sqrt{7x} \)