Question

If a and b are rational number 8 then ,
find the value of a and b
→√3+√2/√3-√2=a+b√6​

Answer 1

$\huge \sf {\orange {\underline {\pink{\underline{Answer :-}}}}}$

• $\bigstar\bold\red{Value\:of\:a=5}$
• $\bigstar\bold\red{Value\:of\:b=2}$

Explanation:

$\Large{\underline{\rm{Given:}}}$

• a and b are rational numbers

$\sf{\dfrac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} } =a+b\sqrt{6} }$

$\Large{\underline{\rm{To\:Find:}}}$

• The value of a and b

$\Large{\underline{\rm{Solution:}}}$

Given,

$\sf{\dfrac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} } =a+b\sqrt{6} }$

Taking the LHS of the equation,

$\sf{\dfrac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} } }$

Rationalising the denominator by multiplying √3 + √2 on both numerator and denominator,

$\sf{\dfrac{\sqrt{3} +\sqrt{2}\times(\sqrt{3} +\sqrt{2} ) }{\sqrt{3} -\sqrt{2}\times(\sqrt{3} +\sqrt{2} ) } }$

Simplifying by using identities:

(a + b)² = a² + b² + 2ab

(a + b) × (a - b) = a² - b²

$\sf \mapsto {\dfrac{3+2+2\sqrt{6} }{3-2} }$

$\sf \mapsto {\dfrac{3+2+2\sqrt{6} }{1} }$

$\sf \mapsto {5+2\sqrt{6} }$

But we know that,

$\sf{\dfrac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} } =a+b\sqrt{6} }$

$\sf \blue{5+2\sqrt{6}=a+b\sqrt{6} }$

Equating it we get,

• a = 5
• b = 2

Hence the value of a is 5 and value of b is 2.

${\huge{\underline{\small{\mathbb{\pink{HOPE\:HELPS\:UH :)}}}}}}$

$\red{\tt{sωєєтcнαям♡~}}$

Answer 2

$\huge \sf {\orange {\underline {\pink{\underline{Answer :-}}}}}$

• $\bigstar\bold\red{Value\:of\:a=5}$
• $\bigstar\bold\red{Value\:of\:b=2}$

Explanation:

$\Large{\underline{\rm{Given:}}}$

a and b are rational numbers

$\sf{\dfrac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} } =a+b\sqrt{6} }$

$\Large{\underline{\rm{To\:Find:}}}$

The value of a and b

$\Large{\underline{\rm{Solution:}}}$

Given,

$\sf{\dfrac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} } =a+b\sqrt{6} }$

Taking the LHS of the equation,

$\sf{\dfrac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} } }$

Rationalising the denominator by multiplying √3 + √2 on both numerator and denominator,

$\sf{\dfrac{\sqrt{3} +\sqrt{2}\times(\sqrt{3} +\sqrt{2} ) }{\sqrt{3} -\sqrt{2}\times(\sqrt{3} +\sqrt{2} ) } }$

Simplifying by using identities:

(a + b)² = a² + b² + 2ab

(a + b) × (a - b) = a² - b²

$\sf \mapsto {\dfrac{3+2+2\sqrt{6} }{3-2} }$

$\sf \mapsto {\dfrac{3+2+2\sqrt{6} }{1} }$

$\sf \mapsto {5+2\sqrt{6} }$

But we know that,

$\sf{\dfrac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} } =a+b\sqrt{6} }$

$\sf \blue{5+2\sqrt{6}=a+b\sqrt{6} }$

Equating it we get,

a = 5

b = 2

Hence the value of a is 5 and value of b is 2.

${\huge{\underline{\small{\mathbb{\pink{HOPE\:HELPS\:UH :)}}}}}}$