Question

If a and b are rational number than find a and b if root 2 + root 3/ 3root 2 - 2 root 3 = a+ b root 6

Answer 1

Answer:

your answer attached in the photo

Answer 2

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Given :

a, b are rational number .

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

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Required to find :

1. Values of a & b ?

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Explanation :

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What is a rationalising factor .

A rationalising factor is a factor of multiplication used to rationalize a surd .

If the product of two surd is a rational number then each surd is the rationalising factor to other .

For example,

$\longrightarrow{\sqrt{2} \times \sqrt{2}}$

$\implies{2}$

Here, √2 (root 2 ) which is a surd is the rationalising factor of each other .

Generally, in most cases we try to rationalize the denominator this is because it helps in solving the question avoiding extra calculations .

In the above question we have to consider the L.H.S. part .

Then , we have to rationalize the denominator .

After, the process of rationalising we have to solve the question further .

However,we will be left with one value .

This the value which must be equivalent with R.H.S. part .

Hence, we can find the values of " a " and " b ".

Knowing this above concept is important to solve questions related to it .

Now, let's crack the answer for the above question .

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Solution :

Given :

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

Step - 1 : Consider the L.H.S part

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

Step - 2 : Rationalising factor

$\small{\small{\small{Rationalising \: \: factor \: \: of \: \: 3\sqrt{2} - 2\sqrt{3} = 3\sqrt{2} + 2\sqrt{3}}}}$

step - 3 : Rationalising the denominator

( Just multiply the rationalising factor with both numerator and denominator )

$\longrightarrow{ \dfrac{\sqrt{2} + \sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} \times {\dfrac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}}}}$

Here use the identity for the denominator since it is in the form of ;

$\boxed{\Rightarrow{(x + y)(x - y) = {x}^{2} - {y}^{2}}}$

So,

$\longrightarrow{ \dfrac{\sqrt{2} + \sqrt{3} ( 3\sqrt{2} + 2\sqrt{3})}{{(3\sqrt{2})}^{2} - {(2\sqrt{3})}^{2}}}$

$\longrightarrow{ \dfrac{\sqrt{2} + \sqrt{3} ( 3\sqrt{2} + 2\sqrt{3})}{(9\times 2) - (4 \times 3)}}$

$\longrightarrow{ \dfrac{\sqrt{2} + \sqrt{3} ( 3\sqrt{2} + 2\sqrt{3})}{18 - 12}}$

$\longrightarrow{ \dfrac{\sqrt{2} + \sqrt{3} ( 3\sqrt{2} + 2\sqrt{3})}{6}}$

Hence, the denominator had got rationalised .

Step - 3 : Now, solve the numerator by multiplying the surds .

$\longrightarrow{ \dfrac{\sqrt{2}(3\sqrt{2}+2\sqrt{3})+\sqrt{3}(3\sqrt{2}+2\sqrt{3})}{6}}$

$\longrightarrow{ \dfrac{3(2)+ 2\sqrt{6} + 3\sqrt{6} + 2(3)}{6}}$

$\longrightarrow{ \dfrac{6+ 2\sqrt{6} + 3\sqrt{6} + 6}{6}}$

Now add up the like terms and numbers .

$\longrightarrow{ \dfrac{12+ 5\sqrt{6}}{6}}$

Step - 4 : Now split this fraction into 2 parts .

$\longrightarrow{ \dfrac{12}{6} + \dfrac{5\sqrt{6}}{6}}$

Step - 5 : Equalling the L.H.S. and R.H.S.

Here the the L.H.S. part is of the form a + b✓6

Hence ,

$\longrightarrow{a = \dfrac{12}{6}}$

$\longrightarrow{b = \dfrac{5}{6}}$

Here, value of a needs to be reduced because ;

According to rational number properties the p,q should be co-primes .

So, reduce the value of a to its simplest form

$\longrightarrow{\dfrac{ \cancel{12}}{ \cancel{6}}}$

$\implies{\dfrac{2}{1} = 2 }$

$\boxed{\longrightarrow{\boxed{Values \: of \: a \: and \: b = 2 \: \: and \: \: \dfrac{5}{6}}}}$

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✅ Hence Solved ..