Question

Simplify: $\begin{array}{l}\frac{4+\sqrt{5}}{4–\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}\end{array}$.

Answer 1

Simplify the expression $\begin{array}{l}\frac{4+\sqrt{5}}{4–\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}\end{array}$

Steps 1: By rationalising $\begin{array}{l}\frac{4+\sqrt{5}}{4–\sqrt{5}}\end{array}$both the numerator and the denominator, we get

$\begin{array}{l}=\frac{(4+\sqrt{5})}{4-\sqrt{5}}\end{array}\times \frac{(4+\sqrt{5})}{(4+\sqrt{5})}$

$=\begin{array}{l}\frac{(4+\sqrt{5}{)}^2}{16-5}\end{array}$

$=\begin{array}{l}\frac{(4+\sqrt{5}{)}^2}{11}\end{array}$

$=\begin{array}{l}\frac{16+5+8\surd 5}{11}\end{array}$ $\left[\because {\left(a+b\right)}^2=a^2+b^2+2\mathrm{ab}\right]$

$=\begin{array}{l}\frac{21+8\surd 5}{11}\end{array}$

Step 2: By rationalising $\begin{array}{l}\frac{4-\sqrt{5}}{4+\sqrt{5}}\end{array}$ both the numerator and the denominator, we get

$\begin{array}{l}\frac{4-\sqrt{5}}{4+\sqrt{5}}\end{array}$$=$$\begin{array}{l}\frac{4-\sqrt{5}}{4+\sqrt{5}}\end{array}\times \frac{4-\sqrt{5}}{4-\sqrt{5}}$

$=\begin{array}{l}\frac{{\left(4-\sqrt{5}\right)}^2}{4^2-{\left(\sqrt{5}\right)}^2}\end{array}$

$=$$\frac{16+5-8\surd 5}{16-5}$

$=$$\frac{21-8\surd 5}{11}$

Step-3: Adding both the expression:

$\begin{array}{l}\frac{4+\sqrt{5}}{4–\sqrt{5}}+\frac{4–\sqrt{5}}{4+\sqrt{5}}\end{array}$$=\begin{array}{l}\frac{21+8\surd 5}{11}\end{array}$$\times$$\frac{21-8\surd 5}{11}$

$=$$\frac{21+8\surd 5+21-8\surd 5}{11}$

$=$$\frac{42}{11}$

Therefore, $\begin{array}{l}\frac{4+\sqrt{5}}{4–\sqrt{5}}+\frac{4–\sqrt{5}}{4+\sqrt{5}}=\frac{42}{11}\end{array}$.