Question

find value of a and b if 3+3root2/3-3root2 = a+b root 2. step by step. explanation needed​

Answer 1

Solution :-

(3 + 3√2)/(3 - 3√2) = a + b√2

Consider LHS

(3 + 3√2)/(3 - 3√2)

Rationalise the denominator i.e multiply both numerator and denominator with (3 + 3√2)

= (3 + 3√2)/(3 - 3√2) × (3 + 3√2)/(3 + 3√2)

= {(3 + 3√2) × (3 + 3√2)}/{(3 - 3√2) × (3 + 3√2)}

= (3 + 3√2)²/{3² - (3√2)²}

Since (x + y)(x - y) = x² - y²

= {3² + (3√2)² + 2(3)(3√2)}/{9 - 9(2)}

Since (a + b)² = a² + b² + 2ab

= {9 + 9(2) + 18√2}/(9 - 18)

= {9 + 18 + 18√2)/(-9)

= (27 + 18√2)/(-9)

= (27/-9) + (18√2/-9)

= (3/-1) + (2√2/-1)

= - 3 + (-2√2)

= - 3 - 2√2

i.e, - 3 - 2√2 = a + b√2

Comparing on both sides

We get, a = - 3

b√2 = - 2√2

b = - 2

Therefore the values of a and b are - 3 and - 2 respectively.

Answer 2

Answer:

L.H.S :

= (3 + 3√2) / (3 - 3√2)

Now , (3 + 3√2) multiply with it numerator and denominator.

= (3+3√2)/(3-3√2) × (3+3√2)/(3+3√2)

= {(3+3√2) × (3+3√2)}/ {(3-3√2) × (3+3√2)}

= (3+3√2)^2 / {3^2 - (3√2^2)}

= (x+y)(x-y) = x^2 - y^2

= {3^2 + (3√2)^2 + 2(3)(3√2)}/{9-9(2)}.

By using formula :

(a+b)^2 = a^2 + b^2 + 2ab.

= {9+9(2) + 18√2}/(9-18).

= {9+18+18√2)/(-9).

= (27+18√2)/(-9).

= (27/-9)+(18√2/-9)

= (3/-1)+2√2/-1)

= -3 + (-2√2)

= -3-2√2

= -3-2√2 = a+b√2

The both sides comparing:

a = -3

b√2 = -2√2

b= -2

The values are :

a = -3.

b= -2.