Question

Rationalise dinometon 3-2√2/3+2√2
नेशनल लाइटर डिनॉमिनेटर 3 - 2 अंडर रूट 2 अपऑन 3 प्लस टू अंडर रूट 2 ​

Answer 1

Given :-

$\implies{ \sf{ \frac{3 - 2 \sqrt{2} }{3 + 2 \sqrt{2} } }}$

We need to rationalise the denominator :

So, we'll multiply ${ \sf{3 - 2 \sqrt{2} }}$ by both the numerator and denominator.

So, it becomes –

$\implies{ \sf{ \frac{3 - 2 \sqrt{2} }{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } }}$

• Identify which is used in numerator :

${ \sf{(a - b) ^{2} = a^{2} - 2ab + b^{2} }}$

• Identify which is used in denominator :

${ \sf{(a + b) \: (a - b) = a ^{2} - b^{2} }}$

$\implies{ \sf{ \frac{(3 - 2 \sqrt{2} )^{2} } {(3)^{2} - (2 \sqrt{2} )^{2} } }}$

$\implies{ \sf{ \frac{(3) ^{2} - 2 \times 3 \times 2 \sqrt{2} + (2 \sqrt{2})^{2} }{9 - 2^{2} \times 2 } }}$

$\implies{ \sf{ \frac{(3) ^{2} - 2 \times 3 \times 2 \sqrt{2} + (2 \sqrt{2})^{2} }{9 - 8 } }}$

$\implies{ \sf{ \frac{9 - 12 \sqrt{2} + 8 }{1}}}$

$\implies{ \sf{17 - 12 \sqrt{2} }}$

Hence, after rationalisation, we get that ${ \sf{7 - 12 \sqrt{2}}}$

Answer 2

Given :-

\implies{ \sf{ \frac{3 - 2 \sqrt{2} }{3 + 2 \sqrt{2} } }}⟹

3+2

2

3−2

2

We need to rationalise the denominator :

So, we'll multiply { \sf{3 - 2 \sqrt{2} }}3−2

2

by both the numerator and denominator.

So, it becomes –

\implies{ \sf{ \frac{3 - 2 \sqrt{2} }{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } }}⟹

3+2

2

3−2

2

×

3−2

2

3−2

2

• Identify which is used in numerator :

{ \sf{(a - b) ^{2} = a^{2} - 2ab + b^{2} }}(a−b)

2

=a

2

−2ab+b

2

• Identify which is used in denominator :

{ \sf{(a + b) \: (a - b) = a ^{2} - b^{2} }}(a+b)(a−b)=a

2

−b

2

\implies{ \sf{ \frac{(3 - 2 \sqrt{2} )^{2} } {(3)^{2} - (2 \sqrt{2} )^{2} } }}⟹

(3)

2

−(2

2

)

2

(3−2

2

)

2

\implies{ \sf{ \frac{(3) ^{2} - 2 \times 3 \times 2 \sqrt{2} + (2 \sqrt{2})^{2} }{9 - 2^{2} \times 2 } }}⟹

9−2

2

×2

(3)

2

−2×3×2

2

+(2

2

)

2

\implies{ \sf{ \frac{(3) ^{2} - 2 \times 3 \times 2 \sqrt{2} + (2 \sqrt{2})^{2} }{9 - 8 } }}⟹

9−8

(3)

2

−2×3×2

2

+(2

2

)

2

\implies{ \sf{ \frac{9 - 12 \sqrt{2} + 8 }{1}}}⟹

1

9−12

2

+8

\implies{ \sf{17 - 12 \sqrt{2} }}⟹17−12

2

Hence, after rationalisation, we get that { \sf{7 - 12 \sqrt{2}}}7−12

2