Question

$\sqrt{27} = 3\sqrt{3}$
$\sqrt{48}$
$\sqrt{63}$
$\sqrt{96}$
$\sqrt{180}$
$\sqrt{x^{11}}$

Explanation:

To simplify each square root, we use the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) where \(a\) is a perfect square.

1. Simplify \(\sqrt{48}\)

Step 1: Factor 48 into perfect square and other factor

We know that \(48 = 16\times3\), where \(16\) is a perfect square.

Step 2: Apply the square - root property

\(\sqrt{48}=\sqrt{16\times3}=\sqrt{16}\times\sqrt{3}\)
Since \(\sqrt{16} = 4\), we have \(\sqrt{48}=4\sqrt{3}\)

2. Simplify \(\sqrt{63}\)

Step 1: Factor 63 into perfect square and other factor

We can write \(63=9\times7\), and \(9\) is a perfect square.

Step 2: Apply the square - root property

\(\sqrt{63}=\sqrt{9\times7}=\sqrt{9}\times\sqrt{7}\)
Since \(\sqrt{9} = 3\), we get \(\sqrt{63}=3\sqrt{7}\)

3. Simplify \(\sqrt{96}\)

Step 1: Factor 96 into perfect square and other factor

We know that \(96 = 16\times6\), with \(16\) being a perfect square.

Step 2: Apply the square - root property

\(\sqrt{96}=\sqrt{16\times6}=\sqrt{16}\times\sqrt{6}\)
Since \(\sqrt{16}=4\), we have \(\sqrt{96} = 4\sqrt{6}\)

4. Simplify \(\sqrt{180}\)

Step 1: Factor 180 into perfect square and other factor

We can factor \(180=36\times5\), and \(36\) is a perfect square.

Step 2: Apply the square - root property

\(\sqrt{180}=\sqrt{36\times5}=\sqrt{36}\times\sqrt{5}\)
Since \(\sqrt{36} = 6\), we get \(\sqrt{180}=6\sqrt{5}\)

5. Simplify \(\sqrt{x^{11}}\)

Step 1: Rewrite the exponent of \(x\)

We can rewrite \(x^{11}=x^{10}\times x\), where \(x^{10}=(x^{5})^{2}\) is a perfect square (assuming \(x\geq0\) for real - valued square roots).

Step 2: Apply the square - root property

\(\sqrt{x^{11}}=\sqrt{x^{10}\times x}=\sqrt{x^{10}}\times\sqrt{x}\)
Since \(\sqrt{x^{10}}=x^{5}\) (for \(x\geq0\)), we have \(\sqrt{x^{11}}=x^{5}\sqrt{x}\)

Final Answers:

• \(\sqrt{48}=\boldsymbol{4\sqrt{3}}\)
• \(\sqrt{63}=\boldsymbol{3\sqrt{7}}\)
• \(\sqrt{96}=\boldsymbol{4\sqrt{6}}\)
• \(\sqrt{180}=\boldsymbol{6\sqrt{5}}\)
• \(\sqrt{x^{11}}=\boldsymbol{x^{5}\sqrt{x}}\) (for \(x\geq0\))

Answer:

To simplify each square root, we use the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) where \(a\) is a perfect square.

1. Simplify \(\sqrt{48}\)

Step 1: Factor 48 into perfect square and other factor

We know that \(48 = 16\times3\), where \(16\) is a perfect square.

Step 2: Apply the square - root property

\(\sqrt{48}=\sqrt{16\times3}=\sqrt{16}\times\sqrt{3}\)
Since \(\sqrt{16} = 4\), we have \(\sqrt{48}=4\sqrt{3}\)

2. Simplify \(\sqrt{63}\)

Step 1: Factor 63 into perfect square and other factor

We can write \(63=9\times7\), and \(9\) is a perfect square.

Step 2: Apply the square - root property

\(\sqrt{63}=\sqrt{9\times7}=\sqrt{9}\times\sqrt{7}\)
Since \(\sqrt{9} = 3\), we get \(\sqrt{63}=3\sqrt{7}\)

3. Simplify \(\sqrt{96}\)

Step 1: Factor 96 into perfect square and other factor

We know that \(96 = 16\times6\), with \(16\) being a perfect square.

Step 2: Apply the square - root property

\(\sqrt{96}=\sqrt{16\times6}=\sqrt{16}\times\sqrt{6}\)
Since \(\sqrt{16}=4\), we have \(\sqrt{96} = 4\sqrt{6}\)

4. Simplify \(\sqrt{180}\)

Step 1: Factor 180 into perfect square and other factor

We can factor \(180=36\times5\), and \(36\) is a perfect square.

Step 2: Apply the square - root property

\(\sqrt{180}=\sqrt{36\times5}=\sqrt{36}\times\sqrt{5}\)
Since \(\sqrt{36} = 6\), we get \(\sqrt{180}=6\sqrt{5}\)

5. Simplify \(\sqrt{x^{11}}\)

Step 1: Rewrite the exponent of \(x\)

We can rewrite \(x^{11}=x^{10}\times x\), where \(x^{10}=(x^{5})^{2}\) is a perfect square (assuming \(x\geq0\) for real - valued square roots).

Step 2: Apply the square - root property

\(\sqrt{x^{11}}=\sqrt{x^{10}\times x}=\sqrt{x^{10}}\times\sqrt{x}\)
Since \(\sqrt{x^{10}}=x^{5}\) (for \(x\geq0\)), we have \(\sqrt{x^{11}}=x^{5}\sqrt{x}\)

Final Answers:

• \(\sqrt{48}=\boldsymbol{4\sqrt{3}}\)
• \(\sqrt{63}=\boldsymbol{3\sqrt{7}}\)
• \(\sqrt{96}=\boldsymbol{4\sqrt{6}}\)
• \(\sqrt{180}=\boldsymbol{6\sqrt{5}}\)
• \(\sqrt{x^{11}}=\boldsymbol{x^{5}\sqrt{x}}\) (for \(x\geq0\))