Question

if a=root 2+1 find (a-1/a)^2
pls answer step by step ​

Answer 1

Step-by-step explanation:

a=√2+1

(a-1/a)²

=(√2+1-1/√2+1)²

=(√2)²/(√2+1)²

=2/(√2)²+1²+2(√2)(1)

=2/2+1+2√2

=2/3+2√2

Answer 2

$\sf \pink{a=\sqrt{2}+1}$

$\sf \dfrac{1}{a}=\dfrac{1}{\sqrt{2}+1}$

On rationalizing the denominator ,

$:\implies \sf \dfrac{1}{a}=\dfrac{1}{\sqrt{2}+1} \times \dfrac{\sqrt{2}-1}{\sqrt{2}-1}$

• (a + b)(a - b) = a² - b²

$:\implies \sf \dfrac{1}{a}=\dfrac{\sqrt{2}-1}{(\sqrt{2})^2-(1)^2}$

$:\implies \sf \dfrac{1}{a}=\dfrac{\sqrt{2}-1}{2-1}$

$:\implies \sf \dfrac{1}{a}=\dfrac{\sqrt{2}-1}{1}$

$:\implies \sf \pink{\dfrac{1}{a}=\sqrt{2}-1}$

So ,

$\mapsto \sf \red{\bigg(a-\dfrac{1}{a} \bigg)^2}$

$\sf \mapsto \bigg( (\sqrt{2}+1) - ( \sqrt{2} -1) \bigg)^2$

$\mapsto \sf \big( \sqrt{2}+1-\sqrt{2}+1 \big)^2$

$\mapsto \sf (2)^2$

$\mapsto \green{4}\ \; \bigstar$