Question

If a and b are rational numbers and √2+2√3/2√2+√3=a+b√6 , find the values of a and b​ please guys answer this for real not for the sake of points

Answer 1

Answer:

a=-10 and b=-5

Step-by-step explanation:

first take LHS

your LHS is √2+2√3/2√2+√3

rationalise the LHS

you get -10-5√6

now LHS is equal to RHS

-10-5√6=a+b√6

hence a=-10  and  b=-5

hope this helps

if any problem then plz let me know

Answer 2

Given :-

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

Required to find :-

• Values of a and b ?

Solution :-

Given data :-

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

we need to find the values of a and b !

So,

Consider the LHS Part

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

Now,

Let's rationalize the denominator of the LHS part ;

Rationalising factor of 2√2 + √3 = 2√2 - √3

Multiply the numerator and denominator with the Rationalising factor

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

Here, we need to use one algebraic Identity ;

That is ;

• ( x + y ) ( x - y ) = x² - y²

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

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

$\tt \dfrac{2 \times \sqrt{ {2}^{2} } - \sqrt{3} \times \sqrt{2} + 2 \times 2 \times \sqrt{2} \times \sqrt{3} \times 2 \sqrt{ {3}^{2} } }{4 \times 2 - 3}$

$\tt \dfrac{2 \times 2 - \sqrt{6} + 4 \sqrt{6} - 2 \times 3 }{8 - 3}$

$\tt \dfrac{4 - \sqrt{6} + 4 \sqrt{6} - 6 }{5}$

$\tt \dfrac{2 + 3 \sqrt{6} }{5}$

This can be written as ;

$\bf\dfrac{2}{5} + \dfrac{3 \sqrt{6} }{5}$

Let's equate both LHS and RHS

$\bf \dfrac{2}{5} + \dfrac{3 \sqrt{6} }{5} = a + b \sqrt{6}$

From the above we can conclude that ;

The LHS is in the form of RHS

So,

$\overline{\downarrow \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \downarrow} \\ \tt{ \dfrac{2}{5} + \frac{3 \sqrt{6} }{5} = a + b \sqrt{6} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline { \uparrow \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \uparrow}$

Therefore,

Value of ' a ' is 2/5

Value of ' b ' is 3/5

Verification :-

It is mentioned that the values of a , b are rational number ;

So,

According to the properties of rational number ;

A number is said to be a rational number if it can be expressed in the form of p/q where p , q are integers . q≠0 and p,q are co - primes

Since,

The values of a , b are satisfying all these conditions hence we say that they are rational numbers .

( The value are expressed in p/q form, where 2 and 5 , 3 and 5 are co - primes similarly the denominator of the values of a ,b are not equal to 0 )

Hence proved !

Additional Information :-

Some of the algebraic identities useful when solving these type of numericals are ;

( x + y ) ( x + y ) = ( x + y )²

( x - y ) ( x - y ) = ( x - y )²

( x - y )² = x² + y² - 2xy

( x + y )² = x² + y² + 2xy

( x + a ) ( x + b ) = x² + x ( a + b ) + ab