Question

A 8root 5 b=8 root5/8-root5 8-root5/8 root5

Answer 1

$a=\dfrac{138}{59}$ and b = 0

Step-by-step explanation:

The complete question:

If $a+8\sqrt{5}b =\dfrac{8+\sqrt{5}}{8-\sqrt{5}} +\dfrac{8-\sqrt{5}}{8+\sqrt{5}}$, find the values of a and b.

We have

$a+8\sqrt{5}b =\dfrac{8+\sqrt{5}}{8-\sqrt{5}} +\dfrac{8-\sqrt{5}}{8+\sqrt{5}}$,

To find, the values of a and b = ?

∴ $a+8\sqrt{5}b =\dfrac{8+\sqrt{5}}{8-\sqrt{5}} +\dfrac{8-\sqrt{5}}{8+\sqrt{5}}$

⇒ $a+8\sqrt{5}b =\dfrac{(8+\sqrt{5})^2+(8+\sqrt{5})^2}{(8+\sqrt{5})(8-\sqrt{5})}$

⇒ $a+8\sqrt{5}b =\dfrac{2(8^2+\sqrt{5}^2)}{8^2-\sqrt{5}^2}$

Using the trigonometric identity,

$a^{2}-b^{2} =(a+b)(a-b)$ and

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

⇒ $a+8\sqrt{5}b =\dfrac{2(64+5)}{64-5}$

⇒ $a+8\sqrt{5}b =\dfrac{138}{59}+0(b)$

Comparing the coefficient of a and b, we get

$a=\dfrac{138}{59}$ and b = 0

Hence, $a=\dfrac{138}{59}$ and b = 0

Answer 2

To find, the values of a and b = ?

∴  a+8root5b =8+root5/8-root5 + 8-root5/8+root5

⇒  a+8root5b =(8+root5)^2 + (8+root5)^2/(8+root5) (8-root5)

⇒  a+8root5b=2(8^2 +root5^2)/8^2-root5^2  

⇒   a+8root5b=2(64+5)/64-5

⇒  a+8root5b=138/59+0(b)

hence A=138/59

B= 0