Question

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Answer 1

Question:-

If $\rm \dfrac{\sqrt{5}+1}{\sqrt{5}-1}+ \dfrac{\sqrt{5}-1}{\sqrt{5}+1}=a+b\sqrt{5},$ find the value of a and b.

Solution:-

Let √5 + 1 = x

and

√5 - 1 = y

Therefore,

$\rm\dashrightarrow \dfrac{x}{y}+ \dfrac{y}{x}=a+b\sqrt{5}$

$\rm\dashrightarrow \dfrac{x^2+y^2}{xy}=a+b\sqrt{5}$

Substituting the values,

$\rm\dashrightarrow \dfrac{(\sqrt{5}+1)^2+(\sqrt{5}-1)^2}{(\sqrt{5}+1)(\sqrt{5}-1)}=a+b\sqrt{5}$

We know that,

• (a + b)(a - b) = a² - b²
• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab

Hence,

$\rm\dashrightarrow \dfrac{(\sqrt{5})^2+(1)^2+2\times \sqrt{5} \times 1+\bigg [ (\sqrt{5})^2+(1)^2-2 \times \sqrt{5}\times 1 \bigg ]}{(\sqrt{5})^2 - (1)^2}=a+b\sqrt{5}$

$\rm\dashrightarrow \dfrac{5 + 1+2 \sqrt{5}+ \bigg [ 5+1-2 \sqrt{5}\bigg ]}{5-1}=a+b\sqrt{5}$

$\rm\dashrightarrow \dfrac{5 + 1+2 \sqrt{5}+ 5+1-2 \sqrt{5}}{4}=a+b\sqrt{5}$

$\rm\dashrightarrow \dfrac{6+2 \sqrt{5}+6-2 \sqrt{5}}{4}=a+b\sqrt{5}$

$\rm\dashrightarrow \dfrac{6+6}{4}=a+b\sqrt{5}$

$\rm\dashrightarrow \dfrac{12}{4}=a+b\sqrt{5}$

$\rm\dashrightarrow 3=a+b\sqrt{5}$

Now 3 can be written as

3 + 0√5

Hence,

$\rm\dashrightarrow 3+0\sqrt{5}=a+b\sqrt{5}$

On comparing we get,

a = 3

b = 0

Answer 2

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