Question

If 1/2+√22+√26=1/√a b^2*[√a+√b- √(a+b)], then values of a and b

Answer 1

Answer:

a = 22 and b = 4

Step-by-step explanation:

Now, 1/(2 + √22 + √26)

= (2 + √22 - √26)/{(2 + √22 + √26)(2 + √22 - √26)}

[ To rationalise the denominator, we multiply both the numerator and the denominator by the conjugate irrational number (2 + √22 - √26) of the denominator (2 + √22 + √26) ]

= (2 + √22 - √26)/{(2 + √22)² - (√26)²}

[ Using the formula (a + b)(a - b) = a² - b² ]

= (2 + √22 - √26)/(4 + 4√22 + 22 - 26)

= (2 + √22 - √26)/(4√22)

Given that,

(2 + √22 - √26)/(4√22) = 1/√(ab²) * {√a + √b - √(a + b)} .....(1)

Now, we rearrange the terms of the left hand side.

So (2 + √22 - √26)/(4√22)

= {√22 + √4 - √(22 + 4)}/√(22 * 4²)

Now comparing with the right hand side of (1), we get

a = 22 and b = 4