Question

What is the purpose of a conjugate?

Answer 1

To simplify radicals (mainly), should I give an example?

Answer 2

Yes please

Answer 3

Lets say you have, \(\LARGE\color{black}{ \frac{1-\sqrt{3}}{4+\sqrt{5}} }\), but you know that you can't have a radical in the denominator, so what you do is that you multiply this by a conjugate.

The concept of a conjugate:
you know that `(a-b)(a+b)` = `a²-b²` and there is no middle term , because it cancels.

THis is why we need it here, to cancel the middle term, will will just have...


\(\LARGE\color{black}{ \frac{(1-\sqrt{3})\color{red}{\times(4-\sqrt{5})}}{(4+\sqrt{5})\color{red}{\times(4-\sqrt{5})}} }\)

(because CONJUNGATE for any a+√b is a-√b )


\(\LARGE\color{black}{ \frac{(1-\sqrt{3})\color{red}{\times(4-\sqrt{5})}}{(4+\sqrt{5})\color{red}{\times(4-\sqrt{5})}} }\)

\(\LARGE\color{black}{ \frac{(1-\sqrt{3})\color{red}{\times(4-\sqrt{5})}}{4^2-\sqrt{5}^2} }\)

Answer 4

The square root of the number that is raised to the second power = nuber 9just like what we have here, with the FIVE)

\(\LARGE\color{black}{ \frac{(1-\sqrt{3})\color{red}{\times(4-\sqrt{5})}}{4^2-\sqrt{5}^2} }\)

\(\LARGE\color{black}{ \frac{(1-\sqrt{3})\color{red}{\times(4-\sqrt{5})}}{4^2-5} }\)

\(\LARGE\color{black}{ \frac{(1-\sqrt{3})\color{red}{\times(4-\sqrt{5})}}{16-5} }\)

\(\LARGE\color{black}{ \frac{(1-\sqrt{3})\color{red}{\times(4-\sqrt{5})}}{11} }\)

Answer 5

So we have rationalized the denominator.