Question

find a and b so that
2+√3/2-√3 = a-b √3
plz answer ​

Answer 1

Given:

The equation -

$\boxed{\bf \leadsto \: \dfrac{2+\sqrt{3}}{2-\sqrt{3}} = a - b\sqrt{3}}$

What To Find:

We have to find -

• The value of a.
• The value of b.

Solution:

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

Rationalise the denominator of LHS with its rationalising factor,

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

Take them as common,

$\sf \leadsto \: \dfrac{(2+\sqrt{3}) \times (2+\sqrt{3})}{(2-\sqrt{3}) \times (2+\sqrt{3})} = a - b\sqrt{3}$

Solve the numerator,

$\sf \leadsto \: \dfrac{2(2+\sqrt{3})+\sqrt{3}(2+\sqrt{3})}{(2-\sqrt{3}) \times (2+\sqrt{3})} = a - b\sqrt{3}$

Multiply the first brackets,

$\sf \leadsto \: \dfrac{(4+2\sqrt{3})+\sqrt{3}(2+\sqrt{3})}{(2-\sqrt{3}) \times (2+\sqrt{3})} = a - b\sqrt{3}$

Multiply the second brackets,

$\sf \leadsto \: \dfrac{(4+2\sqrt{3})+(2\sqrt{3}+ 3)}{(2-\sqrt{3}) \times (2+\sqrt{3})} = a - b\sqrt{3}$

Remove the brackets and rearrange the terms,

$\sf \leadsto \: \dfrac{4+3 +2\sqrt{3}+2\sqrt{3}}{(2-\sqrt{3}) \times (2+\sqrt{3})} = a - b\sqrt{3}$

Solve the terms separately,

$\sf \leadsto \: \dfrac{7+4\sqrt{3}}{(2-\sqrt{3}) \times (2+\sqrt{3})} = a - b\sqrt{3}$

Solve the denominator by using the identity (a - b) (a + b) = a² = b²,

$\sf \leadsto \: \dfrac{7+4\sqrt{3}}{(2)^2-(\sqrt{3})^2} = a - b\sqrt{3}$

Find the squares,

$\sf \leadsto \: \dfrac{7+4\sqrt{3}}{4-3} = a - b\sqrt{3}$

Subtract the numbers,

$\sf \leadsto \: \dfrac{7+4\sqrt{3}}{1} = a - b\sqrt{3}$

Can be written as,

$\sf \leadsto \: 7+4\sqrt{3}= a - b\sqrt{3}$

We can see that -

• $\sf \leadsto \: 7 = a$

• $\sf \leadsto \: 4 = -b \:\:\: \leadsto -4 = b$

Final Answer:

∴ Thus, the value of a is 7 while the value of b is -4.