Question

If a and b are rational numbers find a and b.
√2+√3/3√2-2√2=a+b√6​

Answer 1

Answer:

(3 + sqrt(5)) / (3 - sqrt(5)) = (a + b*sqrt(5)) /1 , we have one equation of 2 fractions.

The cross product gives: (3 + sqrt(5)) = (3 - sqrt(5)) * (a + b*sqrt(5))

Getting rid of the brackets: 3 + sqrt(5) = 3a + 3b*sqrt(5) - 5b - a* sqrt(5)

3 + sqrt(5) = (3a -5b) + (3b - a)*sqrt(5) Separating the Rational from the Irrational,

we get 2 linear equations: 3 = (3a -5b) and

1 = (3b - a)

Multiplying the second equation by 3 gives: 3 = - 3a + 9b

adding the first equation: 3 = 3a -5b

we get : 3+3= - 3a + 3a + 9b - 5b

Eliminating the variable a : 6 = 4 b this gives b = 6/4 = 3/2

Now knowing b=3/2 , we substitute it in 1 = (3b - a) and we get a = 7/2

The answer is: a = 7/2 ; and b = 3/2

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Answer 2

a=1,b=1/2

Step-by-step explanation:

√2+√3/3√2-2√3 first rationalise denominator now we have to multiply and divide denominator =√2+√3/3√2-2√3×3√2+2√3/3√2+2√3

=6+3√6+4+2√6

=10+5√6/(3√2)^2- (2√2)^2

=10+5√6/18-8

=10+5√6/10

=10/10+5√6/10

=1+√6/2=a+b√6

a=1,b=1/2