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,
$\therefore \ \frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}=\frac{21+8\sqrt{5}}{11}=\frac{21-8\sqrt{5}}{11}$
$=\frac{21+8\sqrt{5}+21-8\sqrt{5}}{11}=\frac{42}{11}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
→→
Two chords AB, CD of lengths 5cm, 11cm respectively of a circle are parallel. If the distance between AB and CD is 3cm, find the radius of the circle.
Express in the form of $\frac{\text{p}}{\text{q}}:0.\overline{38}+1.\overline{27}.$
→In figure, $\angle\text{AOC}$ and $\angle\text{BOC}$ from a linear pair. If $a - 2b = 30°$, find $a$ and $b.$ 
→If both $x+1$ and $x-1$ are factors of $a x^3+x^2-2 x+b$, find the values of $a$ and $b$.
→In the adjoining figure, $DE$ is a chord parallel to diameter $AC$ of the circle with centre $O$. If $\angle\text{CBD}=60^\circ,$calculate $\angle\text{CDE}.$ 
→Draw the graph of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes: $2x + y = 6$
→→
The perimeter of a triangular field is $240\ dm$. If two of its sides are $78\ dm$ and $50\ dm,$ find the length of the perpendicular on the side of length $50\ dm$ from the opposite vertex.
→A rhombus shaped sheet with perimeter $40\ cm$ and one diagonal $12\ cm$, is painted on both sides at the rate of Rs. $5$ per $m^2$. Find the cost of painting.
→