Welcome to a lesson on multiplying
radical conjugate expressions.
Radical expressions that involve a sum or difference
of the same two terms are called conjugates.
So here are some examples of radical conjugates.
A plus square root of B, and A minus square root of B,
are radical conjugates, as well as
Square root of A plus square root of B,
and square root of A minus square root of B.
Notice how the terms are the same,
but here we have a sum and here we have a difference,
as well as here we have a sum and here we have a difference.
The product of two conjugates is
the difference of two squares.
And this result is very helpful when
simplifying radical expressions that contain
a sum or difference in the denominator.
Now, there is a short cut to determining
the product of radical conjugates.
If we look at the product of the conjugates
A plus B, and A minus B, this would be equal to
A-squared minus B-squared.
This formula should look familiar.
This is the same formula that we use
to factor a difference of squares.
I'm not a real big fan of the shortcut, notice how
if we multiply these two binomials like we normally would,
we would have four products: one, two, three, four.
Which would give us A-squared, minus AB,
plus AB, minus B-squared.
And notice how the two middle terms are opposites
and therefore, this sum is zero.
So, this simplifies nicely to A-squared minus B-squared.
So, as long as we know how to multiply two binomials,
we don't have to memorize this formula or take the shortcut.
But of course, it is available if you want to use it.
Let's look at an example with radical conjugates.
Notice here we have two plus square root of three,
times two minus square root of three.
So, again, using the technique to multiply two binomials,
we'd have four products, two times two is four,
two times negative square root of three would be
minus two square root of three.
Then we'd have plus square root of three times two
or plus two square root of three, and then we'd have
square root of three times negative square root of three,
which would be minus square root of nine.
Once again, notice how the two middle terms
are opposites, so this sums to zero.
The square root of nine is equal to three, so
this simplifies to four minus three, which equals one.
And if we did use a shortcut, notice that A would be two,
B would be square root of three.
So, we'd have two-squared,
minus square root of three squared,
which would still give us four minus three,
which equals one.
Let's take a look at a couple more examples.
Notice how we do have the product of two conjugates.
Let's first multiply this out like we normally would.
Again, we'd have four products.
One, two, three, four.
So, we have six times six, that's 36.
Six times negative square root of 15,
that's minus six square root of 15.
And then we'd have square root of 15 times six,
so that's plus six square root of 15.
And then we'd have square root of 15 times
negative square root of 15, and that would be
minus the square root of 15 times 15.
Of course, this does simplify, this is a perfect square.
And the two middle terms are opposites.
This sums to zero.
So, we get 36 minus, the square root of 15-squared is 15.
So this product simplifies to 21.
We're going to want to be using the shortcut,
notice A would be 6, B would be square root of 15.
So we'd have six-squared, minus square root of 15 squared.
Which equals 36 minus 15 with the same result.
Either way, the product of the conjugates equals 21.
Let's look at one more example.
Once again, notice how we do have conjugates.
The terms are the same except here we have
a difference and here we have a sum.
Let's first multiply like we normally would.
So we have one, two, three, four products.
Three square root three times three square root three.
You multiply the threes.
Three times three is nine.
Multiply the radicands, two times two.
And then we have, plus
the tough way to think of this as
one square root five.
So we'd have three times one, that's three.
Times the square root of two times five.
And then we'd have negative square root of five,
times three square root of two.
We can put a one here if we want.
So we'd have minus three times
the square root of five times two.
And then finally we'd have minus
the square root of five times five.
So, let's go ahead and simplify.
Notice here we have perfect square factor, two squared.
Here we have a perfect square factor of five squared.
And these won't simplify.
Square root of two-squared is two.
So we have nine times two or 18.
Plus, this would be three, square root ten,
minus three square root ten.
And here we have minus the square root
of five squared, which is five.
Again, these are opposites, so we have 18 minus 5,
which equals 13.
Using the shortcut where A would be three
square root of three, and B would be square root of five.
We would have three, square root two squared,
minus square root of five squared, equals,
well, here at a three, square root of three,
we square the three, that's nine.
And then we square square root of two, that's two,
minus the square root of five squared, is five.
So, 18 minus five of course is 13.
So, as you can see, the shortcut is faster,
but as long as you remember how to multiply binomials,
will always get the correct results.
I hope you found this helpful.