Question

if 6/3√2-2√3=a√2+b√3
find the values of a and b.​

Answer 1

$\huge{\bold{SOLUTION:-}}$

$\frac{6}{3 \sqrt{2} - 2 \sqrt{3} } \times \frac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} }$

$\frac{6(3 \sqrt{2} + 2 \sqrt{3} )}{ ({3 \sqrt{2} })^{2} - ({2 \sqrt{3} })^{2} }$

$\frac{6(3 \sqrt{2} + 2 \sqrt{3}) }{6}$

$3 \sqrt{2} + 2 \sqrt{3}$

ACCORDING TO PROBLEM,

$3 \sqrt{2} + 2 \sqrt{3} = a \sqrt{2} + b \sqrt{3}$

So ,

$\boxed{a = 3}$

$\boxed{b = 2}$

MUST-TO-REMEMBER :-

A surd is a form of irrational number

If n is a positive integer and a is a rational number and is not the nth power of any rational number , then n root a is called a surd

Surd is a number that can't be simplified to remove a square root or cube root etc..

A surd is a square root which cannot be reduced to a whole number

When it is a root and irrational , it is a surd

But all roots are not surds

If two surds are different multiples of same surds , they are similar surds and if they are not multiples they are not similar surds i.e. they are dissimilar surds

By knowing the basic principles , we can solve this type of problems.