Question

Find the value of a and b 2-root5/2+3root5=aroot5+b

Answer 1

Answer:

$\textbf{a = \dfrac{8}{41} and b = - \dfrac{19}{41} }$

Step-by-step explanation:

$\dfrac{2-\sqrt{5}}{2 + 3 \sqrt{5}}\\\\\\= \dfrac{(2-\sqrt{5})(2 - 3 \sqrt{5})}{(2 + 3 \sqrt{5})(2 - 3 \sqrt{5})}\\\\\\= \dfrac{4 - 6 \sqrt{5} - 2 \sqrt{5} + 3 (\sqrt{5})^2}{(2)^2 - (3 \sqrt{5})^2}\\\\\\= \dfrac{4 - 8 \sqrt{5} + 3(5)}{4 - 9(5)}\\\\\\= \dfrac{4 + 15 - 8 \sqrt{5}}{4 - 45}\\\\\\= \dfrac{19 - 8 \sqrt{5}}{-41}\\\\\\= \dfrac{-}{-} \times \dfrac{19 - 8 \sqrt{5}}{-41}\\\\\\= \dfrac{8 \sqrt{5} - 19}{41}\\\\\\= \dfrac{8\sqrt{5}}{41} - \dfrac{19}{41}$

$= \dfrac{8 \sqrt{5}}{41} + (- \dfrac{19}{41})\\\\\\\text{Now, This is equal to a \sqrt{5} + b }\\\\\\\bold{\therefore \: \textbf{a = \dfrac{8}{41} and b = - \dfrac{19}{41} }}$

Answer 2

value of a= 8

41

value of b= 19

41