Question

$\sqrt{5 } - 1 \div \sqrt{5} + 1 + \sqrt{5 } + 1 \div \sqrt{5} - 1 = a + b \sqrt{5}$
answer
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Answer 1

Answer:

$\sf{The \ value \ of \ a \ is \ 3 \ and \ value}$

$\sf{b \ is \ zero.}$

Given:

$\sf{a+b=\dfrac{\sqrt5-1}{\sqrt5+1}+\dfrac{\sqrt5+1}{\sqrt5-1}}$

To find:

• The value of a and b.

Solution:

$\sf{a+b=\dfrac{\sqrt5-1}{\sqrt5+1}+\dfrac{\sqrt5+1}{\sqrt5-1}}$

$\sf{=\dfrac{(\sqrt5-1)^{2}+(\sqrt5+1)^{2}}{(\sqrt5+1)(\sqrt5-1)}}$

$\sf{=\dfrac{2(\sqrt5)^{2}+2(1)^{2}}{(\sqrt5)^{2}-1^{2}}}$

$\sf{=\dfrac{10+2}{4}}$

$\sf{=3}$

$\sf{\therefore{a+b\sqrt5=3}}$

$\sf{\therefore{a=3 \ and \ b=0}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ a \ is \ 3 \ and \ value}}}$

$\sf\purple{\tt{b \ is \ zero.}}$