Question

Find the value of a and b if 2-√5/2+3√5=√5a+b

Answer 1

Answer:

Step-by-step explanation:

The given equation is:

$\frac{2-\sqrt{5}}{2+3\sqrt{5}}=\sqrt{5}a+b$

Taking the LHS of the above equation, we get

$\frac{2-\sqrt{5}}{2+3\sqrt{5}}=\frac{2-\sqrt{5}}{2+3\sqrt{5}}{\times}\frac{2-3\sqrt{5}}{2-3\sqrt{5}}$

=$\frac{4-6\sqrt{5}-2\sqrt{5}+15}{4-45}$

=$\frac{19-7\sqrt{5}}{-41}$

=$\frac{-7\sqrt{5}+19}{-41}$

Now, comparing with RHS, we have

$a=\frac{7}{41}$ and $b=\frac{-19}{41}$

Answer 2

Given:

$\frac{2-\sqrt{5}}{2+3\sqrt{5}}=\sqrt{5} a+b$

To find:

The value of a and b

Solution:

$\frac{2-\sqrt{5}}{2+3\sqrt{5}}$

By using rationalization

$\frac{(2-\sqrt{5})\times (2-3\sqrt{5})}{(3\sqrt{5}+2)(2-3\sqrt{5})}$

$\frac{4-6\sqrt{5}-2\sqrt{5}+15}{-45+4}$

$\frac{-8\sqrt{5}+19}{-41}$

$\frac{8\sqrt{5}-19}{41}$

$\frac{8}{41}\sqrt{5}-\frac{19}{41}$

By comparing we get

$a=\frac{8}{41},b=-\frac{19}{41}$