Question
Simplify: $\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$
✓
Answer
For rationalising the denominator of a number, we multilply its numerator and denominator by its rationalising factor.
If $a$ and $b$ are integers, then $\big(\text{a}+\sqrt{\text{b}}\big)$ and $\big(\text{a}-\sqrt{\text{b}}\big)$ are rationalising factor of each other, as $\big(\text{a}+\sqrt{\text{b}}\big)\big(\text{a}-\sqrt{\text{b}}\big)=\big(\text{a}^2-\text{b}\big),$ which is rational. Let us rationalise the denominator of the first term on the left hand side.
We have, $\frac{4+\sqrt{5}}{4-\sqrt{5}}=\frac{4+\sqrt{5}}{4-\sqrt{5}}\times\frac{4+\sqrt{5}}{4+\sqrt{5}}$
$=\frac{\big(4+\sqrt{5}\big)^2}{(4)^2-\big(\sqrt{5}\big)^2}$
$=\frac{(4)^2+2(4)\big(\sqrt{5}\big)+\big(\sqrt{5}\big)^2}{16-5}$
$\frac{4+\sqrt{5}}{4-\sqrt{5}}=\frac{16+8\sqrt{5}+5}{11}=\frac{21+8\sqrt{5}}{11} \ ...(1)$
Now consider the denominator of the second term on the left hand side:
$\frac{4-\sqrt{5}}{4+\sqrt{5}}=\frac{4-\sqrt{5}}{4+\sqrt{5}}\times\frac{4-\sqrt{5}}{4-\sqrt{5}}$
$=\frac{\big(4-\sqrt{5}\big)^2}{(4)^2-\big(\sqrt{5}\big)^2}$
$=\frac{(4)^2-2(4)\big(\sqrt{5}\big)+\big(\sqrt{5}\big)^2}{16-5}$
$\frac{4-\sqrt{5}}{4+\sqrt{5}}=\frac{16-8\sqrt{5}+5}{11}=\frac{21-8\sqrt{5}}{11} \ ...(2)$
Adding equations $(1)$ and $(2)$, we have,
$\therefore \ \frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}=\frac{21+8\sqrt{5}}{11}=\frac{21-8\sqrt{5}}{11}$
$=\frac{21+8\sqrt{5}+21-8\sqrt{5}}{11}=\frac{42}{11}$
If $a$ and $b$ are integers, then $\big(\text{a}+\sqrt{\text{b}}\big)$ and $\big(\text{a}-\sqrt{\text{b}}\big)$ are rationalising factor of each other, as $\big(\text{a}+\sqrt{\text{b}}\big)\big(\text{a}-\sqrt{\text{b}}\big)=\big(\text{a}^2-\text{b}\big),$ which is rational. Let us rationalise the denominator of the first term on the left hand side.
We have, $\frac{4+\sqrt{5}}{4-\sqrt{5}}=\frac{4+\sqrt{5}}{4-\sqrt{5}}\times\frac{4+\sqrt{5}}{4+\sqrt{5}}$
$=\frac{\big(4+\sqrt{5}\big)^2}{(4)^2-\big(\sqrt{5}\big)^2}$
$=\frac{(4)^2+2(4)\big(\sqrt{5}\big)+\big(\sqrt{5}\big)^2}{16-5}$
$\frac{4+\sqrt{5}}{4-\sqrt{5}}=\frac{16+8\sqrt{5}+5}{11}=\frac{21+8\sqrt{5}}{11} \ ...(1)$
Now consider the denominator of the second term on the left hand side:
$\frac{4-\sqrt{5}}{4+\sqrt{5}}=\frac{4-\sqrt{5}}{4+\sqrt{5}}\times\frac{4-\sqrt{5}}{4-\sqrt{5}}$
$=\frac{\big(4-\sqrt{5}\big)^2}{(4)^2-\big(\sqrt{5}\big)^2}$
$=\frac{(4)^2-2(4)\big(\sqrt{5}\big)+\big(\sqrt{5}\big)^2}{16-5}$
$\frac{4-\sqrt{5}}{4+\sqrt{5}}=\frac{16-8\sqrt{5}+5}{11}=\frac{21-8\sqrt{5}}{11} \ ...(2)$
Adding equations $(1)$ and $(2)$, we have,
$\therefore \ \frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}=\frac{21+8\sqrt{5}}{11}=\frac{21-8\sqrt{5}}{11}$
$=\frac{21+8\sqrt{5}+21-8\sqrt{5}}{11}=\frac{42}{11}$
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