Question

write a factor that you can use to rationalize the denominator of \\(\frac{3m}{sqrt4{4}}\\). a rationalizing factor is \\(square\\)

Explanation:

Step1: Recall the rule for rationalizing a fourth root denominator.

To rationalize a denominator with a fourth root, we need to make the radicand a perfect fourth power. The denominator here is \(\sqrt[4]{4}\), and \(4 = 2^2\). We need to multiply by a factor that will make the exponent of 2 in the radicand a multiple of 4.

Step2: Determine the missing exponent.

The current exponent of 2 in the radicand is 2. We need to get it to 4 (the index of the root). So we need to multiply by \(2^{4 - 2}=2^2 = 4\) inside the fourth root? Wait, no, more accurately, for \(\sqrt[4]{a}\), to rationalize, we multiply numerator and denominator by \(\sqrt[4]{a^{n}}\) where \(n\) is such that \(a\times a^{n}\) is a perfect fourth power. Here, \(a = 4=2^2\), so we need to multiply by \(\sqrt[4]{2^{2}}\) (wait, no, let's do it properly). The general formula for rationalizing \(\sqrt[n]{x}\) is to multiply by \(\sqrt[n]{x^{n - k}}\) where \(k\) is the exponent of \(x\) modulo \(n\). Wait, maybe simpler: the denominator is \(\sqrt[4]{4}=\sqrt[4]{2^2}\). To make the radicand a perfect fourth power, we need to multiply by \(\sqrt[4]{2^2}\) because \((2^2)\times(2^2)=2^4\), which is a perfect fourth power. Wait, no: \(\sqrt[4]{2^2}\times\sqrt[4]{2^2}=\sqrt[4]{2^4}=2\). Wait, actually, if we have \(\sqrt[4]{4}=\sqrt[4]{2^2}\), then to rationalize, we multiply numerator and denominator by \(\sqrt[4]{2^2}\) (because \(2^2\times2^2 = 2^4\), so the fourth root of that is 2, a rational number). Wait, let's check: \(\frac{3m}{\sqrt[4]{4}}\times\frac{\sqrt[4]{4^3}}{\sqrt[4]{4^3}}\)? No, wait, no. Wait, the index is 4, and the radicand is 4, which is \(2^2\). The exponent of 2 in the radicand is 2. We need the exponent to be a multiple of 4. So we need to multiply by \(2^{4 - 2}=2^2 = 4\) inside the fourth root? Wait, no, the radicand is 4, so we can write \(\sqrt[4]{4}=\sqrt[4]{2^2}\). To make the radicand a perfect fourth power, we need to multiply by \(\sqrt[4]{2^2}\) (since \(2^2\times2^2 = 2^4\)). So the rationalizing factor is \(\sqrt[4]{4}\)? Wait, no, let's do it step by step. Let's denote the denominator as \(d=\sqrt[4]{4}=\sqrt[4]{2^2}\). We want to find a factor \(f\) such that \(d\times f\) is rational (i.e., a perfect fourth power times a rational number, but actually, we want \(d\times f\) to be an integer or a rational number without a radical). So \(d\times f=\sqrt[4]{2^2}\times f\). We need \(\sqrt[4]{2^2}\times f\) to be a perfect fourth power. Let \(f = \sqrt[4]{2^2}\), then \(\sqrt[4]{2^2}\times\sqrt[4]{2^2}=\sqrt[4]{2^4}=2\), which is rational. Wait, but \(2^2\times2^2 = 2^4\), so yes. Alternatively, since \(4 = 2^2\), the fourth root of 4 is \(2^{2/4}=2^{1/2}\), but we need to rationalize the fourth root. Wait, maybe another approach: the denominator is \(\sqrt[4]{4}\), so to make the radicand a perfect fourth power, we need to multiply by \(\sqrt[4]{4^3}\)? Wait, no, let's recall that for a radical of index \(n\), \(\sqrt[n]{a^m}\), to rationalize, we multiply by \(\sqrt[n]{a^{n - m}}\) so that \(a^m\times a^{n - m}=a^n\), a perfect \(n\)-th power. Here, \(n = 4\), \(a = 2\), \(m = 2\) (since \(4 = 2^2\)). So \(n - m=4 - 2 = 2\). Therefore, we multiply by \(\sqrt[4]{2^2}=\sqrt[4]{4}\). Wait, but let's check: \(\sqrt[4]{4}\times\sqrt[4]{4}=\sqrt[4]{16}=\sqrt[4]{2^4}=2\), which is rational. So the rationalizing factor is \(\sqrt[4]{4}\) or \(\sqrt[4]{2^2}\), but we can also write 4 as \(2^2\), so maybe \(\sqrt[4]{4}\) is the same as \(\sqrt[4]{2^2}\). Alternatively, since \(4 = 2^2\), we can write the denominator as \(\sqrt[4]{2^2}\), and to make the exponent of 2 a multiple of 4, we need to multiply by \(\sqrt[4]{2^2}\) (because \(2^2\times2^2 = 2^4\)). So the rationalizing factor is \(\sqrt[4]{4}\) (or \(\sqrt[4]{2^2}\), which is the same as \(\sqrt[4]{4}\)). Wait, but let's verify with the original expression. If we multiply numerator and denominator by \(\sqrt[4]{4}\), we get:

Numerator: \(3m\times\sqrt[4]{4}\)

Denominator: \(\sqrt[4]{4}\times\sqrt[4]{4}=\sqrt[4]{16}=2\)

So the denominator becomes 2, which is rational. Therefore, the rationalizing factor is \(\sqrt[4]{4}\) (or \(\sqrt[4]{2^2}\), or we can simplify \(\sqrt[4]{4}\) as \(4^{1/4}=2^{2/4}=2^{1/2}=\sqrt{2}\), but wait, no, \(\sqrt[4]{4}\) is \(4^{1/4}\), and \(\sqrt{2}\) is \(2^{1/2}=4^{1/4}\), so they are the same. Wait, but let's check the exponent: \(4 = 2^2\), so \(\sqrt[4]{4}=\sqrt[4]{2^2}=2^{2/4}=2^{1/2}=\sqrt{2}\). Wait, but if we multiply \(\sqrt[4]{4}\) by \(\sqrt[4]{4}\), we get \(\sqrt[4]{16}=\sqrt[4]{2^4}=2\), which is correct. Alternatively, maybe the rationalizing factor is \(\sqrt[4]{4^3}\)? Wait, no, let's do the exponent correctly. The index is 4, so we need the radicand to be a perfect fourth power. The current radicand is 4, which is \(2^2\). We need to multiply by \(2^{4 - 2}=2^2 = 4\) inside the fourth root? Wait, no, the radicand is 4, so to make it a perfect fourth power, we need to multiply by \(4^{3}\) inside the fourth root? Wait, no, that's not right. Wait, the formula for rationalizing \(\sqrt[n]{a}\) is to multiply numerator and denominator by \(\sqrt[n]{a^{n - 1}}\) if \(a\) is not a perfect \(n\)-th power. Wait, no, for example, to rationalize \(\sqrt{2}\), we multiply by \(\sqrt{2}\) (index 2, so \(n - 1 = 1\), so \(\sqrt{2^1}\)). For \(\sqrt[3]{2}\), we multiply by \(\sqrt[3]{2^2}\) (index 3, \(n - 1 = 2\)). So for index 4, we multiply by \(\sqrt[4]{a^{3}}\) where \(a\) is the radicand. Wait, in our case, the radicand is 4, so \(a = 4\), so we multiply by \(\sqrt[4]{4^3}\). Let's check: \(\sqrt[4]{4}\times\sqrt[4]{4^3}=\sqrt[4]{4\times4^3}=\sqrt[4]{4^4}=4\), which is rational. Oh! That's a better way. Because \(4\times4^3 = 4^4\), which is a perfect fourth power. So the rationalizing factor is \(\sqrt[4]{4^3}\) or \(\sqrt[4]{64}\) (since \(4^3 = 64\)). Wait, let's verify:

Denominator: \(\sqrt[4]{4}\times\sqrt[4]{4^3}=\sqrt[4]{4\times4^3}=\sqrt[4]{4^4}=4\), which is rational.

Ah, I see, my earlier mistake was not using the correct exponent. For a radical of index \(n\), to rationalize \(\sqrt[n]{a}\), we multiply by \(\sqrt[n]{a^{n - 1}}\) so that \(a\times a^{n - 1}=a^n\), a perfect \(n\)-th power. So here, \(n = 4\), \(a = 4\), so we multiply by \(\sqrt[4]{4^{4 - 1}}=\sqrt[4]{4^3}\).

Let's check with exponents: \(\sqrt[4]{4}=4^{1/4}\), \(\sqrt[4]{4^3}=4^{3/4}\). Multiplying them gives \(4^{1/4 + 3/4}=4^1 = 4\), which is rational. Perfect.

So the rationalizing factor is \(\sqrt[4]{4^3}\) or \(\sqrt[4]{64}\) (since \(4^3 = 64\)).

Answer:

\(\sqrt[4]{4^3}\) (or \(\sqrt[4]{64}\))