Given:
The equation is:
[tex]\left(3+4\sqrt{3}\right)\left(2-a\sqrt{3}\right)=-18+2\sqrt{3}[/tex]
To find:
The value of a.
Solution:
We have,
[tex]\left(3+4\sqrt{3}\right)\left(2-a\sqrt{3}\right)=-18+2\sqrt{3}[/tex]
On simplification, we get
[tex](3)(2)+(4\sqrt{3})(2)+(3)(-a\sqrt{3})+(4\sqrt{3})(-a\sqrt{3})=-18+2\sqrt{3}[/tex]
[tex]6+8\sqrt{3}-3a\sqrt{3}-4a(3)=-18+2\sqrt{3}[/tex]
[tex]6+8\sqrt{3}-3a\sqrt{3}-12a=-18+2\sqrt{3}[/tex]
[tex](6-12a)+(8-3a)\sqrt{3}=-18+2\sqrt{3}[/tex]
On comparing both sides, we get
[tex]6-12a=-18[/tex]
[tex]-12a=-18-6[/tex]
[tex]a=\dfrac{-24}{-12}[/tex]
[tex]a=2[/tex]
And,
[tex]8-3a=2[/tex]
[tex]-3a=2-8[/tex]
[tex]a=\dfrac{-6}{-3}[/tex]
[tex]a=2[/tex]
Therefore, the value of a is 2.
