Question

pls tell me the answer and explain me how to solve it ​

Answer 1

Answer:

a=7,b=4

Step-by-step explanation:

2+root3/2-root3

rationalise numerator and denominator with 2+root3

Answer 2

Answer:

Required value of a is 7 and b is 4.

Step-by-step explanation:

Given,

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

On rationalizing LHS : { Dividing and multiplying by the conjugate of denominator }

$\implies \dfrac{2 + \sqrt{3} }{2 - \sqrt{3} } = a + b \sqrt{3} \\ \\ \\ \implies \dfrac{2 + \sqrt{3} }{ 2 - \sqrt{3} } \times \dfrac{2 + \sqrt{3} }{2 + \sqrt{3} } = a + b \sqrt{3} \\ \\ \\ \implies \frac{(2 + \sqrt{3} ) {}^{2} }{(2 - \sqrt{3})(2 + \sqrt{3} )} = a + b \sqrt{3}$

Using ( a + b )^2 = a^2 + b^2 + 2ab, numerator of LHS is now 4 + 3 + 4√3 = 7 + 4√3

Using ( a - b )( a + b ) = a^2 - b^2, denominator of LHS is now 4 - 3 = 1

= > ( 7 + 4√3 ) / 1 = a + b√3

Rational number on LHS = Rational number on RHS = 7 = a

Irrational number on LHS = Irrational number on RHS = 4 = b

*Conjugate of a number refers to the same number with opposite sign of irrational number in that number.