Question

simplifying radicals
review

• simplify a square root that is not a perfect square by rewriting it as a product of perfect squares and other factors.

\\(\sqrt{20} = \sqrt{4 \cdot 5} = \sqrt{4} \cdot \sqrt{5} = 2\sqrt{5}\\)

• simplify a square root with variables by rewriting the expression to show the perfect square factors. then remove them in the radicand.

\\(\sqrt{45a^5} = \sqrt{9 \cdot 5 \cdot a^2 \cdot a^2 \cdot a} \leftarrow \text{rewrite } a^5 \text{ as } a^2 \cdot a^2 \cdot a.\\)
\\(\quad \quad \quad = \sqrt{3 \cdot 3} \cdot \sqrt{5} \cdot \sqrt{a^2 \cdot a^2} \cdot \sqrt{a}\\)
\\(\quad \quad \quad = 3a^2\sqrt{5a}\\)

simplify \\(\sqrt{2} \cdot \sqrt{12x^3}\\).
step 1 rewrite the second radical to show the perfect square factors.
\\(\sqrt{2} \cdot \sqrt{12x^3}\\)
\\(= \sqrt{2} \cdot \sqrt{4 \cdot \square \cdot x^2 \cdot \square}\\)
\\(= \sqrt{2} \cdot \sqrt{4} \cdot \sqrt{3} \cdot \square \cdot \square\\)

step 2 simplify the perfect squares.
\\(\sqrt{2} \cdot \sqrt{4} \cdot \sqrt{3} \cdot \sqrt{x^2} \cdot \sqrt{x}\\)
\\(= \sqrt{2} \cdot \square \cdot \sqrt{3} \cdot \square \cdot \sqrt{x}\\)

step 3 multiply the radicals.
\\(\sqrt{2} \cdot 2 \cdot \sqrt{3} \cdot x \cdot \sqrt{x}\\)
\\(= 2x \cdot \sqrt{\square \cdot \square \cdot \square}\\)
\\(= 2x \cdot \sqrt{\square}\\)

\\(\sqrt{2} \cdot \sqrt{12x^3} = 2x\sqrt{6x}\\)

Explanation:

Step 1: Rewrite the second radical

We know that \(12 = 4\times3\) and \(x^{3}=x^{2}\times x\). So we rewrite \(\sqrt{12x^{3}}\) as \(\sqrt{4\times3\times x^{2}\times x}\). Then, using the property of square roots \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (\(a\geq0,b\geq0\)), we can split the square root:
\(\sqrt{2}\cdot\sqrt{12x^{3}}=\sqrt{2}\cdot\sqrt{4\times3\times x^{2}\times x}=\sqrt{2}\cdot\sqrt{4}\cdot\sqrt{3}\cdot\sqrt{x^{2}}\cdot\sqrt{x}\)

Step 2: Simplify perfect squares

We know that \(\sqrt{4} = 2\) and \(\sqrt{x^{2}}=x\) (assuming \(x\geq0\)). So substituting these values in:
\(\sqrt{2}\cdot\sqrt{4}\cdot\sqrt{3}\cdot\sqrt{x^{2}}\cdot\sqrt{x}=\sqrt{2}\cdot2\cdot\sqrt{3}\cdot x\cdot\sqrt{x}\)

Step 3: Multiply the radicals

First, multiply the non - radical terms and the radical terms separately. The non - radical terms are \(2\) and \(x\), and the radical terms are \(\sqrt{2}\), \(\sqrt{3}\), and \(\sqrt{x}\). Using the property \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\), we have \(\sqrt{2}\cdot\sqrt{3}\cdot\sqrt{x}=\sqrt{2\times3\times x}=\sqrt{6x}\). And the non - radical terms multiply to \(2x\). So:
\(\sqrt{2}\cdot2\cdot\sqrt{3}\cdot x\cdot\sqrt{x}=2x\cdot\sqrt{2\times3\times x}=2x\sqrt{6x}\)

Answer:

\(2x\sqrt{6x}\)