Question

rewrite the following expression in simplest radical form given ( x > 0 ).
( sqrt4{16x^2} cdot sqrt3{216x^{11}} )

Explanation:

Step1: Simplify the fourth root

Simplify $\sqrt[4]{16x^2}$. We know that $16 = 2^4$, so $\sqrt[4]{16x^2}=\sqrt[4]{2^4x^2}=2x^{\frac{2}{4}} = 2x^{\frac{1}{2}}$.

Step2: Simplify the cube root

Simplify $\sqrt[3]{216x^{11}}$. We know that $216 = 6^3$, so $\sqrt[3]{216x^{11}}=\sqrt[3]{6^3x^{11}}$. Now, we can rewrite $x^{11}$ as $x^{9 + 2}=x^{9}\cdot x^{2}=(x^{3})^{3}\cdot x^{2}$. Then $\sqrt[3]{6^3x^{11}} = 6x^{3}\cdot x^{\frac{2}{3}}=6x^{3+\frac{2}{3}}=6x^{\frac{11}{3}}$. Wait, actually, more accurately, when simplifying $\sqrt[3]{x^{11}}$, we divide the exponent by 3: $11\div3 = 3$ with a remainder of 2. So $x^{11}=x^{3\times3 + 2}=(x^{3})^{3}\cdot x^{2}$, so $\sqrt[3]{x^{11}}=x^{3}\sqrt[3]{x^{2}}$. So $\sqrt[3]{216x^{11}}=6x^{3}\sqrt[3]{x^{2}}$.

Wait, maybe a better way: Let's handle the exponents properly. Recall that $\sqrt[n]{a^m}=a^{\frac{m}{n}}$. So for $\sqrt[4]{16x^2}$, $16 = 2^4$, so $\sqrt[4]{2^4x^2}=2x^{\frac{2}{4}}=2x^{\frac{1}{2}}$. For $\sqrt[3]{216x^{11}}$, $216 = 6^3$, so $\sqrt[3]{6^3x^{11}} = 6x^{\frac{11}{3}}$. Now, we can rewrite $x^{\frac{11}{3}}$ as $x^{3+\frac{2}{3}}=x^{3}\cdot x^{\frac{2}{3}}$, but maybe we can also express the original product first.

Wait, actually, let's do it step by step. Let's first simplify each radical:

$\sqrt[4]{16x^2}$:
Since $16 = 2^4$, so $\sqrt[4]{16x^2} = \sqrt[4]{2^4 \cdot x^2} = 2 \cdot x^{\frac{2}{4}} = 2x^{\frac{1}{2}}$ (because $\sqrt[4]{a^4}=a$ and $\sqrt[4]{x^2}=x^{\frac{2}{4}}$).

$\sqrt[3]{216x^{11}}$:
Since $216 = 6^3$, so $\sqrt[3]{216x^{11}}=\sqrt[3]{6^3 \cdot x^{11}} = 6 \cdot x^{\frac{11}{3}}$. Now, let's rewrite $x^{\frac{11}{3}}$ as $x^{3 + \frac{2}{3}} = x^3 \cdot x^{\frac{2}{3}}$, so $\sqrt[3]{216x^{11}} = 6x^3 \cdot x^{\frac{2}{3}} = 6x^3\sqrt[3]{x^2}$ (since $x^{\frac{2}{3}}=\sqrt[3]{x^2}$).

Now, multiply the two simplified radicals:

$(2x^{\frac{1}{2}}) \cdot (6x^3\sqrt[3]{x^2})$

First, multiply the coefficients: $2 \times 6 = 12$.

Then, multiply the $x$ terms: $x^{\frac{1}{2}} \times x^3 = x^{\frac{1}{2} + 3} = x^{\frac{1}{2} + \frac{6}{2}} = x^{\frac{7}{2}}$.

So now we have $12x^{\frac{7}{2}}\sqrt[3]{x^2}$. But this seems complicated. Wait, maybe I made a mistake in the approach. Let's try another way. Let's express both radicals with exponents and then multiply, then convert back to radical form.

First, $\sqrt[4]{16x^2} = (16x^2)^{\frac{1}{4}} = (2^4x^2)^{\frac{1}{4}} = 2^{4\times\frac{1}{4}}x^{2\times\frac{1}{4}} = 2x^{\frac{1}{2}}$.

Second, $\sqrt[3]{216x^{11}} = (216x^{11})^{\frac{1}{3}} = (6^3x^{11})^{\frac{1}{3}} = 6^{3\times\frac{1}{3}}x^{11\times\frac{1}{3}} = 6x^{\frac{11}{3}}$.

Now, multiply the two expressions:

$(2x^{\frac{1}{2}}) \times (6x^{\frac{11}{3}}) = 2\times6\times x^{\frac{1}{2}+\frac{11}{3}} = 12x^{\frac{3}{6}+\frac{22}{6}} = 12x^{\frac{25}{6}}$.

Now, let's convert $x^{\frac{25}{6}}$ back to radical form. $\frac{25}{6}=4 + \frac{1}{6}$? Wait, no, $\frac{25}{6}=4\frac{1}{6}$? Wait, $6\times4 = 24$, so $25 = 24 + 1$, so $\frac{25}{6}=4 + \frac{1}{6}$? No, $25\div6 = 4$ with a remainder of 1? Wait, no, $6\times4 = 24$, so $25 = 24 + 1$, so $\frac{25}{6}=4 + \frac{1}{6}$? Wait, no, $x^{\frac{25}{6}} = x^{4 + \frac{1}{6}} = x^4 \cdot x^{\frac{1}{6}} = x^4\sqrt[6]{x}$. Wait, no, $\frac{25}{6}= \frac{24 + 1}{6}=4 + \frac{1}{6}$, so $x^{\frac{25}{6}}=x^4 \cdot x^{\frac{1}{6}}=\sqrt[6]{x}\cdot x^4$. Wait, but maybe we can express the exponent as a mixed number for the radical. Wait, actually, $x^{\frac{25}{6}} = x^{\frac{24 + 1}{6}} = x^{\frac{24}{6}+\frac{1}{6}} = x^4 \cdot x^{\frac{1}{6}} = x^4\sqrt[6]{x}$.

Wait, but let's check the multiplication again. Wait, $\frac{1}{2} + \frac{11}{3} = \frac{3}{6} + \frac{22}{6} = \frac{25}{6}$, that's correct. So $12x^{\frac{25}{6}} = 12x^4 \cdot x^{\frac{1}{6}} = 12x^4\sqrt[6]{x}$. But maybe there's a better way. Wait, perhaps I made a mistake in simplifying the radicals initially. Let's try simplifying the radicals first before multiplying.

Wait, let's re-express the original expression:

$\sqrt[4]{16x^2} \cdot \sqrt[3]{216x^{11}}$

First, simplify $\sqrt[4]{16x^2}$:

$16 = 2^4$, so $\sqrt[4]{16x^2} = 2\sqrt[4]{x^2} = 2x^{\frac{2}{4}} = 2x^{\frac{1}{2}}$ (since $x > 0$, we don't have to worry about absolute value).

Simplify $\sqrt[3]{216x^{11}}$:

$216 = 6^3$, so $\sqrt[3]{216x^{11}} = 6\sqrt[3]{x^{11}}$. Now, $x^{11} = x^{9 + 2} = (x^3)^3 \cdot x^2$, so $\sqrt[3]{x^{11}} = x^3\sqrt[3]{x^2}$. Therefore, $\sqrt[3]{216x^{11}} = 6x^3\sqrt[3]{x^2}$.

Now, multiply the two simplified radicals:

$2x^{\frac{1}{2}} \cdot 6x^3\sqrt[3]{x^2} = 12x^{\frac{1}{2} + 3}\sqrt[3]{x^2} = 12x^{\frac{7}{2}}\sqrt[3]{x^2}$.

Now, let's express $x^{\frac{7}{2}}$ as $x^3\sqrt{x}$ (since $\frac{7}{2} = 3 + \frac{1}{2}$, so $x^{\frac{7}{2}} = x^3 \cdot x^{\frac{1}{2}} = x^3\sqrt{x}$). So now we have:

$12x^3\sqrt{x} \cdot \sqrt[3]{x^2}$

To combine these radicals, we can express them with a common index. The least common multiple of 2 and 3 is 6. So:

$\sqrt{x} = x^{\frac{1}{2}} = x^{\frac{3}{6}} = \sqrt[6]{x^3}$

$\sqrt[3]{x^2} = x^{\frac{2}{3}} = x^{\frac{4}{6}} = \sqrt[6]{x^4}$

Therefore, $\sqrt{x} \cdot \sqrt[3]{x^2} = \sqrt[6]{x^3} \cdot \sqrt[6]{x^4} = \sqrt[6]{x^3 \cdot x^4} = \sqrt[6]{x^7} = \sqrt[6]{x^6 \cdot x} = x\sqrt[6]{x}$

So now, going back to the expression:

$12x^3 \cdot x\sqrt[6]{x} = 12x^{4}\sqrt[6]{x}$

Wait, that's the same as before. So $12x^4\sqrt[6]{x}$. Let's check if this is correct.

Alternatively, let's verify with exponents:

$x^{\frac{7}{2}} \cdot x^{\frac{2}{3}} = x^{\frac{7}{2} + \frac{2}{3}} = x^{\frac{21 + 4}{6}} = x^{\frac{25}{6}}$, which is the same as $x^4 \cdot x^{\frac{1}{6}}$ (since $4 = \frac{24}{6}$, so $\frac{24}{6} + \frac{1}{6} = \frac{25}{6}$), so $x^{\frac{25}{6}} = x^4 \cdot x^{\frac{1}{6}} = x^4\sqrt[6]{x}$. Then multiplying by 12 gives $12x^4\sqrt[6]{x}$.

Let me check with a specific value of $x$. Let's take $x = 64$ (a perfect sixth power, $64 = 2^6$).

First, compute the original expression:

$\sqrt[4]{16 \times 64^2} \cdot \sqrt[3]{216 \times 64^{11}}$

Compute $16 \times 64^2 = 16 \times 4096 = 65536$. $\sqrt[4]{65536} = 16$ (since $16^4 = 65536$).

Compute $216 \times 64^{11}$. $64 = 2^6$, so $64^{11} = 2^{66}$. $216 = 6^3 = 2^3 \times 3^3$. So $216 \times 64^{11} = 2^3 \times 3^3 \times 2^{66} = 2^{69} \times 3^3$. $\sqrt[3]{2^{69} \times 3^3} = 2^{23} \times 3$.

Now, multiply the two results: $16 \times 2^{23} \times 3 = 2^4 \times 2^{23} \times 3 = 2^{27} \times 3$.

Now, compute the simplified expression $12x^4\sqrt[6]{x}$ with $x = 64$:

$x = 64 = 2^6$, so $x^4 = (2^6)^4 = 2^{24}$, $\sqrt[6]{x} = \sqrt[6]{2^6} = 2$. So $12x^4\sqrt[6]{x} = 12 \times 2^{24} \times 2 = 12 \times 2^{25} = 3 \times 4 \times 2^{25} = 3 \times 2^{27}$, which matches the original expression's result. So that's correct.

So the simplest radical form is $12x^4\sqrt[6]{x}$.

Answer:

$12x^{4}\sqrt[6]{x}$