Question

5. Find the values of a and b if:
6/ 3 root 2 + 2 root 3
= a√2+b/3​

Answer 1

Step-by-step explanation:

6/(3√2-2√3)

Here we have to rationalise the denominator,

Rationalising factor=(3√2+2√3)

=6/(3√2-2√3)×(3√2+2√3)/(3√2+2√3)

=6(3√2+2√3)/(3√2-2√3)(3√2+2√3)

=(18√2+12√3)/[(3√2)²-(2√3)²]

=(18√2+12√3)/(9(2)-4(3))

=(18√2+12√3)/18-12

=(18√2+12√3)/6

=(18√2/6)+(12√3/6)

=3√2+2√3

=3√2-(-2√3)

Compare this with 3√2-a√3

3√2-(-2√3)=3√2-a√3

Therefore, a= -2

Hope it helps...

Answer 2

Answer:

LHS = ( √2 + √3 )/(3√2-2√3)

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

=[6+2√6+3√6+6]/[(3√2)²-(2√3)²]

= [12+5√6]/[18-12]

= ( 12 + 5√6 )/6

= 12/6 + 5√6/6

= 2 + (5/6)√6 ---( 1 )

= a + b√6-------( 2 )