Question

find a and b if 7+√5/3+√5-7-√5/3-√5=a+√5b tell yrrr​

Answer 1

Answer:

a=0 and b=-2

Step-by-step explanation:

$\frac{7+\sqrt5}{3+\sqrt5}$

$=\frac{7+\sqrt5}{3+\sqrt5}*\frac{3-\sqrt5}{3-\sqrt5}$

$=\frac{(7+\sqrt5)(3-\sqrt5)}{3^2-{\sqrt5}^2}$

$=\frac{21-7\sqrt5+3\sqrt5-5}{9-5}$

$=\frac{16-4\sqrt5}{4}$

$=\frac{4(4-\sqrt5)}{4}$

$\frac{7+\sqrt5}{3+\sqrt5}=4-\sqrt5$...................(1)

$\frac{7-\sqrt5}{3-\sqrt5}$

$=\frac{7-\sqrt5}{3-\sqrt5}*\frac{3+\sqrt5}{3+\sqrt5}$

$=\frac{(7-\sqrt5)(3+\sqrt5)}{3^2-{\sqrt5}^2}$

$=\frac{21+7\sqrt5-3\sqrt5-5}{9-5}$

$=\frac{16+4\sqrt5}{4}$

$=\frac{4(4+\sqrt5)}{4}$

$\frac{7-\sqrt5}{3-\sqrt5}=4+\sqrt5$................(2)

(1)-(2) gives

$\frac{7+\sqrt5}{3+\sqrt5}-\frac{7-\sqrt5}{3-\sqrt5}=(4-\sqrt5)-(4+\sqrt5)$

$\implies\:\frac{7+\sqrt5}{3+\sqrt5}-\frac{7-\sqrt5}{3-\sqrt5}=4-\sqrt5-4-\sqrt5$

$\implies\:\frac{7+\sqrt5}{3+\sqrt5}-\frac{7-\sqrt5}{3-\sqrt5}=0-2\sqrt5$

comparing with

$\implies\:\frac{7+\sqrt5}{3+\sqrt5}-\frac{7-\sqrt5}{3-\sqrt5}=a+b\sqrt5$

we get

a=0 and b=-2