Question

3Find the value of a in the equation,(3+4√3)(2-a√3)=-18+2√3​

Answer 1

Given:

The equation is:

$\left(3+4\sqrt{3}\right)\left(2-a\sqrt{3}\right)=-18+2\sqrt{3}$

To find:

The value of a.

Solution:

We have,

$\left(3+4\sqrt{3}\right)\left(2-a\sqrt{3}\right)=-18+2\sqrt{3}$

On simplification, we get

$(3)(2)+(4\sqrt{3})(2)+(3)(-a\sqrt{3})+(4\sqrt{3})(-a\sqrt{3})=-18+2\sqrt{3}$

$6+8\sqrt{3}-3a\sqrt{3}-4a(3)=-18+2\sqrt{3}$

$6+8\sqrt{3}-3a\sqrt{3}-12a=-18+2\sqrt{3}$

$(6-12a)+(8-3a)\sqrt{3}=-18+2\sqrt{3}$

On comparing both sides, we get

$6-12a=-18$

$-12a=-18-6$

$a=\dfrac{-24}{-12}$

$a=2$

And,

$8-3a=2$

$-3a=2-8$

$a=\dfrac{-6}{-3}$

$a=2$

Therefore, the value of a is 2.