Question

if 2/(√3+√5) + 5(√3-√5)=a√3+b√5, find a and b​

Answer 1

Answer: I am just considering that your question is 2/(√3+√5) + 5/(√3-√5) = a√3 + b√5.

I just guess you missed the division sign after 5.

First if we see LHS

Given, 2/(√3+√5) +5(√3-√5)

So we have to rationalise the denominator

Then we get,

                      [2(√3-√5) / (√3)²-(√5)²] +[5 (√3+√5) / (√3)²-(√5)²]

                   [ 2(√3-√5) /-2] +  [5√3 + 5√5 /-2]

So 2 and -2 get's cancelled, and the result becomes

          -√3+√5 +[ -5√3 - 5√5 /2]...................[changing the signs from denominator to numerator]                              

⇒ -2√3+ 2√5 - 5√3 - 5√5 ÷ 2...................[taking LCM 2 and multiplying it to -√3 and √5]

⇒ -7√3 - 3√5 ÷2

∴  -7√3 /2 - 3√5 /2 .......[LHS complete]

Now in RHS, it is given that,

a√3+b√5

So we will equate the first part in the LHS with the first in RHS.

∴ -7√3/2 = a√3

⇒ -7/2 = a

Also, we will equate the second part of LHS with the second part of RHS.

∴ -3√5 /2 = b√5

⇒ -3/2 = b

Thus the required value of a is -7/2 or -3.5, and for b is -3/2 or -1.5

Hope my answer helped :)