Question

How to solve this?
Find the values of a and b:$(2-3\sqrt{6}) /(5+\sqrt{6})= a+b\sqrt{6}$

Answer 1

Proper Question

Find the values of rational numbers $a$ and $b$, where $\dfrac{2-3\sqrt{6} }{5+\sqrt{6} } =a+\sqrt{6} b$.

Main Concept

• Rationalizing the denominator

If square roots are in the denominator, we need to square each of them. The removal of square roots is called rationalization.

The appropriate identity here is $(a+b)(a-b)=a^2-b^2$, so we are going to multiply the terms with different signs to square both terms.

Solution

Let's rationalize the denominator by suitable identity.

Used identity:-

• $(a+b)(a-b)=a^2-b^2$

Rationalization:-

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

$=\dfrac{(2-3\sqrt{6} )(5-\sqrt{6} )}{(5+\sqrt{6} )(5-\sqrt{6} )}$

$=\dfrac{10-2\sqrt{6} -15\sqrt{6} +18}{25-6}$

$=\dfrac{28-17\sqrt{6} }{19}$

$=\dfrac{28}{19} -\dfrac{17}{19} \sqrt{6}$

Comparing it we get,

$a+\sqrt{6} b=\dfrac{28}{19} -\dfrac{17}{19} \sqrt{6}$

$\therefore a=\dfrac{28}{19} ,b=-\dfrac{17}{19}$

Answer 2

Given :-

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

To Find :-

Value of a and b

Solution :-

By rationalizing

$\sf \dfrac{2-3\sqrt{6}}{5+\sqrt{6}} \times \dfrac{5-\sqrt{6}}{5-\sqrt{6}}$

$\sf \dfrac{2-3\sqrt{6} \times 5 - \sqrt{6}}{5 + \sqrt{6}\times 5 - \sqrt{6}}$

$\sf\dfrac{10 - (2\sqrt{6}) -(15\sqrt{6})+18}{25-6}$

$\sf\dfrac{28 - (2\sqrt{6}) -(15\sqrt{6})}{25-6}$

$\sf\dfrac{28-2\sqrt{6}-15\sqrt{6}}{19}$

$\sf\dfrac{28-17\sqrt{6}}{19}$

Value of a = 28/19

Value of b = -(17/19)