Question

If a = √5 +1/√5 -1 and b= √5-1/√5+1, then find a^2+ab+b^2/a^2-ab+b^2.

Answer 1

Given:

$a=\frac{\sqrt{5}+1}{\sqrt{5}-1}\;\;\&\;\;b=\frac{\sqrt{5}-1}{\sqrt{5}+1}$

$a+b=\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}$

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

$a+b=\frac{5+1+2\sqrt{5}+5+1-2\sqrt{5}}{\sqrt{5}^2-1^2}$

$a+b=\frac{12}{4}$

$\implies\bf\,a+b=3$

$a-b=\frac{\sqrt{5}+1}{\sqrt{5}-1}-\frac{\sqrt{5}-1}{\sqrt{5}+1}$

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

$a-b=\frac{5+1+2\sqrt{5}-5-1+2\sqrt{5}}{\sqrt{5}^2-1^2}$

$a-b=\frac{4\sqrt{5}}{4}$

$\implies\bf\,a-b=\sqrt5$

$ab=\frac{\sqrt{5}+1}{\sqrt{5}-1}{\times}\frac{\sqrt{5}-1}{\sqrt{5}+1}=1$

$\implies\bf\,ab=1$

Now,

$\frac{a^2+ab+b^2}{a^2-ab+b^2}$

$=\frac{(a+b)^2-ab}{(a-b)^2+ab}$

$=\frac{(3)^2-1}{(\sqrt5)^2+1}$

$=\frac{9-1}{5+1}$

$=\frac{8}{6}$

$=\frac{4}{3}$

$\implies\boxed{\bf\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{4}{3}}$

Find more:

If x=√5+1÷√5-1 and y=√5-1÷√5+1 find the value of x^2+ xy +y^2

https://brainly.in/question/3728106

Answer 2

Answer:

xyz

Step-by-step explanation:

Given:

Now,

Find more:

If x=√5+1÷√5-1 and y=√5-1÷√5+1 find the value of x^2+ xy +y^2