Question

6+root2/6-root2- 6-root2/6+root2=a-broot2

Answer 1

ANSWER:

Given:

• (6+√2)/(6-√2) - (6-√2)/(6+√2) = a - b√2

To Find:

• Value of a and b

Solution:

We are given that,

$\implies\dfrac{6+\sqrt2}{6-\sqrt2}-\dfrac{6-\sqrt2}{6+\sqrt2}=a-b\sqrt2$

Taking LCM,

$\implies\dfrac{(6+\sqrt)^2-(6-\sqrt2)^2}{(6-\sqrt2)(6+\sqrt2)}=a-b\sqrt2$

We know that,

$\hookrightarrow x^2-y^2=(x+y)(x-y)$

So,

$\implies\dfrac{(6+\sqrt)^2-(6-\sqrt2)^2}{(6-\sqrt2)(6+\sqrt2)}=a-b\sqrt2$

$\implies\dfrac{[(6+\sqrt)+(6-\sqrt2)][(6+\sqrt2)-(6-\sqrt2)]}{(6-\sqrt2)(6+\sqrt2)}=a-b\sqrt2$

$\implies\dfrac{[6+\sqrt+6-\sqrt2][6+\sqrt2-6+\sqrt2]}{(6-\sqrt2)(6+\sqrt2)}=a-b\sqrt2$

$\implies\dfrac{(12)(2\sqrt2)}{(6-\sqrt2)(6+\sqrt2)}=a-b\sqrt2$

We know that,

$\hookrightarrow (x+y)(x-y)=x^2-y^2$

So,

$\implies\dfrac{24\sqrt2}{(6)^2-(\sqrt2)^2}=a-b\sqrt2$

$\implies\dfrac{24\sqrt2}{36-2}=a-b\sqrt2$

$\implies\dfrac{24\sqrt2}{34}=a-b\sqrt2$

$\implies\dfrac{12\sqrt2}{17}=a-b\sqrt2$

$\implies0+\dfrac{12\sqrt2}{17}=a-b\sqrt2$

On comparing the terms,

$\implies\bf a=0,\:\:b=-\dfrac{12}{17}$

Answer 2

Answer:

a=0, b=-12/17

HOPE IT HELPS!