Question

How to rationalise denominator of
5+2 root 3 by 7 +4root3=a-6root of 3

Answer 1

rationalization
5+2√3÷7+4√3=a-6√3 
multiply 5+2√3÷7+4√3 by 7-4√3÷7-4√3
5+2√3÷7+4√3×7-4√3÷7-4√3=a-6√3
(5+2√3)×(7-4√3)÷(7+4√3)×(7-4√3)=a-6√3
(35-20√3+14√3-24)÷(49-3×16)=a-6√3
(11-6√3)÷(49-48)=a-6√3
(11-6√3)÷1=a-6√3
11-6√3=a-6√3
11=a
∴a=11