Question

Identify the conjugate of...

Answer 1

\[\LARGE 3-\sqrt{2}\]

Answer 2

\[\large A) 3-\sqrt{2} \]
\[\large B) 3+\sqrt{2}\]
\[\large C) 2+\sqrt{3}\]
\[\large D) \sqrt{2}-3\]

Answer 3

What's a conjugate?

Answer 4

A conjugate means you reverse the sign on the term with the square root, but leave the other term the same.

Answer 5

Here's why you want the conjugate...

You usually don't want to leave numbers with square roots in the denominator of fractions. You can eliminate the square root by multiplying both numerator and denominator by the conjugate.

so if you had 3 + sqrt(2) in the denominator, and you multiplied by 3 - sqrt(2), you would get:

( 3 + sqrt(2) ) (3 - sqrt(2) ) = (3)(3) + 3sqrt(2) - 3sqrt(2) - sqrt(2)sqrt(2)

and that simplifies to just 9 - 2 = 7.

Answer 6

So, much simpler version:

if you have:\[a + \sqrt{b}\]
Then the conjugate is the same but with a minus in front of the sqrt...
\[a - \sqrt{b}\]

Answer 7

So the answer would be B?

Answer 8

By multiplying something by its "conjugate", you rationalize it.

Something under a square root is [in most cases] irrational. To "rationlize" that expression would mean to get rid of the square root.

\[\large (a+\sqrt{b})(a-\sqrt{b}) \\ \large = a^2 -b\]

Answer 9

Notice that \(a^2 - b\) is rational whereas the original expression \(a+\sqrt{b}\) is not.

To make that rational, you had to multiply it by its conjugate, \(a-\sqrt{b}\).

Answer 10

B is the correct answer.

Answer 11

Thank you both :D

Answer 12

Thanks @muntoo - I had not thought about how the word "rationalize" was linked to this (although it seems obvious now).

Actually, I wasn't sure the term "conjugate" even applies to this... I think I learned it as rationalizing. Then, with complex numbers, we used the term "conjugate", and the process is so similar that it makes sense to use it even in problems like this.