Question

1. rationalize each numerator.

a. \\(\frac{\sqrt{5} - 1}{4}\\)
b. \\(\frac{2 - 3\sqrt{2}}{2}\\)
c. \\(\frac{\sqrt{5} + 2}{2\sqrt{5} - 1}\\)

Explanation:

Part (a)

Step1: Identify the conjugate of the numerator

The numerator is \(\sqrt{5} - 1\), so its conjugate is \(\sqrt{5} + 1\). We multiply the numerator and the denominator by this conjugate.
\[
\frac{\sqrt{5} - 1}{4} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1}
\]

Step2: Multiply the numerators and denominators

Using the difference of squares formula \((a - b)(a + b)=a^{2}-b^{2}\) for the numerator:
\[
\text{Numerator}: (\sqrt{5}-1)(\sqrt{5} + 1)=(\sqrt{5})^{2}-1^{2}=5 - 1 = 4
\]
\[
\text{Denominator}:4\times(\sqrt{5}+1)=4\sqrt{5}+4
\]
So the fraction becomes \(\frac{4}{4\sqrt{5}+4}\)

Step3: Simplify the fraction

We can factor out a 4 from the denominator:
\[
\frac{4}{4(\sqrt{5}+1)}=\frac{1}{\sqrt{5}+1}
\]
We can also rationalize the denominator further by multiplying numerator and denominator by \(\sqrt{5}-1\) (the conjugate of \(\sqrt{5} + 1\)):
\[
\frac{1}{\sqrt{5}+1}\times\frac{\sqrt{5}-1}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{(\sqrt{5})^{2}-1^{2}}=\frac{\sqrt{5}-1}{5 - 1}=\frac{\sqrt{5}-1}{4}? \text{Wait, no, we made a mistake. Wait, actually, when we rationalize the numerator, we just want to eliminate the radical from the numerator. Wait, the original step of multiplying by the conjugate of the numerator is correct. Wait, let's re - do step 2.}
\]
Wait, the formula for \((a - b)(a + b)=a^{2}-b^{2}\), so \((\sqrt{5}-1)(\sqrt{5}+1)=5 - 1=4\), denominator is \(4\times(\sqrt{5}+1)\). Then \(\frac{4}{4(\sqrt{5}+1)}=\frac{1}{\sqrt{5}+1}\) is correct. But if we want to have the numerator without radicals, we have achieved that because the numerator is 1 now. Wait, the original numerator was \(\sqrt{5}-1\), after multiplying by \(\sqrt{5}+1\), the numerator is 4 (no radicals). So the rationalized form (with numerator rationalized) is \(\frac{4}{4(\sqrt{5}+1)}=\frac{1}{\sqrt{5}+1}\) or we can simplify by dividing numerator and denominator by 4: \(\frac{1}{\sqrt{5}+1}\)

Part (b)

Step1: Identify the conjugate of the numerator

The numerator is \(2-3\sqrt{2}\), its conjugate is \(2 + 3\sqrt{2}\). Multiply numerator and denominator by \(2 + 3\sqrt{2}\)
\[
\frac{2-3\sqrt{2}}{2}\times\frac{2 + 3\sqrt{2}}{2 + 3\sqrt{2}}
\]

Step2: Multiply the numerators and denominators

Using the difference of squares formula \((a - b)(a + b)=a^{2}-b^{2}\) for the numerator:
\[
\text{Numerator}: (2-3\sqrt{2})(2 + 3\sqrt{2})=2^{2}-(3\sqrt{2})^{2}=4-9\times2=4 - 18=- 14
\]
\[
\text{Denominator}:2\times(2 + 3\sqrt{2}) = 4+6\sqrt{2}
\]
So the fraction becomes \(\frac{-14}{4 + 6\sqrt{2}}\)

Step3: Simplify the fraction

We can factor out a 2 from the denominator:
\[
\frac{-14}{2(2 + 3\sqrt{2})}=\frac{-7}{2 + 3\sqrt{2}}
\]

Part (c)

Step1: Identify the conjugate of the numerator

The numerator is \(\sqrt{5}+2\), its conjugate is \(\sqrt{5}-2\). Multiply the numerator and the denominator by \(\sqrt{5}-2\)
\[
\frac{\sqrt{5}+2}{2\sqrt{5}-1}\times\frac{\sqrt{5}-2}{\sqrt{5}-2}
\]

Step2: Multiply the numerators and denominators

For the numerator, use the difference of squares formula \((a + b)(a - b)=a^{2}-b^{2}\):
\[
\text{Numerator}:(\sqrt{5}+2)(\sqrt{5}-2)=(\sqrt{5})^{2}-2^{2}=5 - 4 = 1
\]
For the denominator, use the distributive property \((a + b)(c + d)=ac+ad+bc+bd\):
\[

$$\begin{align*} (2\sqrt{5}-1)(\sqrt{5}-2)&=2\sqrt{5}\times\sqrt{5}-2\sqrt{5}\times2-1\times\sqrt{5}+1\times2\\ &=2\times5-4\sqrt{5}-\sqrt{5}+2\\ &=10 + 2-(4\sqrt{5}+\sqrt{5})\\ &=12-5\sqrt{5} \end{align*}$$

\]
So the fraction becomes \(\frac{1}{12 - 5\sqrt{5}}\)
We can also rationalize the denominator by multiplying numerator and denominator by the conjugate of the denominator \(12 + 5\sqrt{5}\):
\[
\frac{1}{12-5\sqrt{5}}\times\frac{12 + 5\sqrt{5}}{12 + 5\sqrt{5}}=\frac{12 + 5\sqrt{5}}{12^{2}-(5\sqrt{5})^{2}}=\frac{12 + 5\sqrt{5}}{144-125}=\frac{12 + 5\sqrt{5}}{19}
\]
But since we are only asked to rationalize the numerator, the form \(\frac{1}{12 - 5\sqrt{5}}\) (with numerator rationalized) is also correct.

Final Answers:

a. \(\frac{1}{\sqrt{5}+1}\) (or \(\frac{\sqrt{5}-1}{4}\) if we simplify the denominator - rationalized form, but the numerator - rationalized form is \(\frac{4}{4\sqrt{5}+4}=\frac{1}{\sqrt{5}+1}\))
b. \(\frac{-7}{2 + 3\sqrt{2}}\)
c. \(\frac{1}{12 - 5\sqrt{5}}\) (or \(\frac{12 + 5\sqrt{5}}{19}\) if we rationalize the denominator as well)

Answer:

Step1: Identify the conjugate of the numerator

The numerator is \(\sqrt{5}+2\), its conjugate is \(\sqrt{5}-2\). Multiply the numerator and the denominator by \(\sqrt{5}-2\)
\[
\frac{\sqrt{5}+2}{2\sqrt{5}-1}\times\frac{\sqrt{5}-2}{\sqrt{5}-2}
\]

Step2: Multiply the numerators and denominators

For the numerator, use the difference of squares formula \((a + b)(a - b)=a^{2}-b^{2}\):
\[
\text{Numerator}:(\sqrt{5}+2)(\sqrt{5}-2)=(\sqrt{5})^{2}-2^{2}=5 - 4 = 1
\]
For the denominator, use the distributive property \((a + b)(c + d)=ac+ad+bc+bd\):
\[

$$\begin{align*} (2\sqrt{5}-1)(\sqrt{5}-2)&=2\sqrt{5}\times\sqrt{5}-2\sqrt{5}\times2-1\times\sqrt{5}+1\times2\\ &=2\times5-4\sqrt{5}-\sqrt{5}+2\\ &=10 + 2-(4\sqrt{5}+\sqrt{5})\\ &=12-5\sqrt{5} \end{align*}$$

\]
So the fraction becomes \(\frac{1}{12 - 5\sqrt{5}}\)
We can also rationalize the denominator by multiplying numerator and denominator by the conjugate of the denominator \(12 + 5\sqrt{5}\):
\[
\frac{1}{12-5\sqrt{5}}\times\frac{12 + 5\sqrt{5}}{12 + 5\sqrt{5}}=\frac{12 + 5\sqrt{5}}{12^{2}-(5\sqrt{5})^{2}}=\frac{12 + 5\sqrt{5}}{144-125}=\frac{12 + 5\sqrt{5}}{19}
\]
But since we are only asked to rationalize the numerator, the form \(\frac{1}{12 - 5\sqrt{5}}\) (with numerator rationalized) is also correct.

Final Answers:

a. \(\frac{1}{\sqrt{5}+1}\) (or \(\frac{\sqrt{5}-1}{4}\) if we simplify the denominator - rationalized form, but the numerator - rationalized form is \(\frac{4}{4\sqrt{5}+4}=\frac{1}{\sqrt{5}+1}\))
b. \(\frac{-7}{2 + 3\sqrt{2}}\)
c. \(\frac{1}{12 - 5\sqrt{5}}\) (or \(\frac{12 + 5\sqrt{5}}{19}\) if we rationalize the denominator as well)