Question

2+√3/2-√3=a+b√3 this is the question

Answer 1

We have:

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

We have to find the value of a and b.

Solution:

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

Rationalising numerator and denominator, we get

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

Using the trogonometric identity:

(a + b)(a - b) = $a^2-b^2$

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

Using the trogonometric identity:

$(a+b)^2=a^2+2ab+b^2$

⇒ $\dfrac{4+4\sqrt{3} +3 }{4-3}=a+b\sqrt{3}$

⇒ $7+4\sqrt{3}=a+b\sqrt{3}$

Comparing both sides, we get

a = 7 and b = 4

∴ a = 7 and b = 4