Question

If (√2 + √3)/(3√2 - 2√3) = a - b√6, then find the value of a and b.

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Answer 1

Step-by-step explanation:

Given:-

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

To Find:-

• Find the value of a and b

Solution:-

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

By rationalizing the denominator:-

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

$\sf \longrightarrow \dfrac{( \sqrt{2} + \sqrt{3}) \times ( 3 \sqrt{2} + 2 \sqrt{3}) }{(3 \sqrt{2} - 2 \sqrt{3} ) \times (3 \sqrt{2} + 2\sqrt{3}) } = a - b \sqrt{6}$

$\sf \longrightarrow \dfrac{\sqrt{2} (3 \sqrt{2} ) + \sqrt{2} (2 \sqrt{3} ) + \sqrt{3}( 3 \sqrt{2} ) + \sqrt{3}( 2 \sqrt{3}) }{(3 \sqrt{2} ) ^{2} - (2 \sqrt{3} )^{2} } = a - b \sqrt{6}$

$\sf \longrightarrow \dfrac{6 + 2 \sqrt{6} + 3 \sqrt{6} + 6 }{18 - 12 } = a - b \sqrt{6}$

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

$\sf \longrightarrow \dfrac{12 + 5\sqrt{6} }{6 } = a - b \sqrt{6}$

$\sf \longrightarrow \: \dfrac{12}{6} + \dfrac{ 5\sqrt{6} }{6 } = a - b \sqrt{6}$

$\sf \longrightarrow 2+ \dfrac{ 5\sqrt{6} }{6 } = a - b \sqrt{6}$

$\sf Comparing \: \: 2+ \dfrac{ 5\sqrt{6} }{6 } \: \: with \: \: a - b \sqrt{6}$

$\dashrightarrow \underline{\boxed{ \bf a = 2 \: , \: b = \frac{ - 5}{6}} }$

Answer 2

Data :

(√2 + √3)/(3√2 - 2√3) = a - b√6

Solving :

= ( √2 + √3 / 3 √2 - 2 √3 ) × ( 3 √2 + 2 √3 / 3 √2 + z √3 ) = a - b √6

= [ √2 ( 3 √2 ) + √2 ( 2 √3 ) + √3 ( 3 √2 ) + √3 ( 2 √3 ) ] / [ ( 3 √2 )^2 - ( 2 √3 )^2 ] = a - b √6

= ( 6 + 2 √6 + 3 √6 + 6 ) / ( 18 - 12 ) = a - b √6

= ( 6 + 6 + 2 √6 + 3 √6 ) / 6 = a - b √6

= ( 12 + 5 √6 ) / 6 = a - b √6

= 2 + 5 √6 / 6 = a - b √6

= Form = a - b √6

So, a = 2

a = 2 b = -5 / 6