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# Worked example: rationalizing the denominator

Sal rationalizes the denominator of the expression (16+2x²)/(√8). Created by Sal Khan and Monterey Institute for Technology and Education.

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• In the last simplification, when everything is being simplified by two, why is the eight outside the square root (this eight->8 square root of 8) not simplified to a four?
• He didn't do that because the equation (8 * sqrt(2) + sqrt(2) * x^2) / 2 can be considered simplified. However, you could actually divide the 8 by 2 and get 4 * sqrt(2) + sqrt(2)/2 * x^2.
• why does he write √2x^2 and not x^2√2? (which to me looks like he wrote (√2)(x^2) instead of x^2√2)

im mildly confused why because i see x^2 as a whole number so is it just a matter of preference similar to writing (√2)(2) instead of 2√2 ?

i hope my question makes sense
• Technically, you could write it the way Sal did, but it is much less confusing to write x^2sqrt2 so that is how it is usually written.
• I was asked this exercise during a challenge test: Rationalize the denominator of this expression:
(1+3√x)/(√(x-3)-1)

Is there anyone who kindly would explain how to rationalize this expression's denominator?
• Sure. Let us look at the denominator only:
It has the pattern of (√a) − b
We multiply this by (√a) + b
[(√a) - b][ (√a) + b]
Using the FOIL method you probably learned toward the end of Algebra I:
(√a)(√a) + b√a - b√a - b(b)
= a - b²
Of course, to be able to do this, we have to multiply the numerator by (√a) + b, so that we don't change the value of the expression.
So, for your problem we would do the following
Denominator:
[ √(x−3) − 1 ][ √(x−3) + 1]
= (x−3) - 1
= x-4
But we also have to multiply the numerator by √(x-3) + 1]
(1+3√x)[√(x-3) + 1]
You can multiply that out if you want, but it can't really be simplified.
So the numerator is: (1+3√x)[√(x-3) + 1]
The denominator is: (x−4)
• What If there was a sum of two radicals? For example 1÷(√3 + √2)?
• Then to rationalize the denominator, you would multiply by the conjugate of the denominator over itself. The conjugate of a binomial has the same first term and the opposite second term. So you would multiply by (sqrt(3) - sqrt(2)) / (sqrt(3) - sqrt(2))
• why does he say principle square root of two instead of just the square root of two?
• The principle square root basically means the positive square root of two, as the square root could be positive or negative. Generally, when a problem asks you to find the square root of something or has some kind of radical, it means the principle square root. Note I said "generally".
• At about Sal multiplies the fraction by the square root of 2 over the square root of 2. Isn't that the same as multiplying by one?
• Yes, it is exactly the same as multiplying by one. And since anything multiplied by one is unchanged, you can multiply by one any time you like. If you multiply by √2 / √2 then you don't change the value of the expression, but you introduce a factor of √2 on both the top and bottom of the fraction -- and this is helpful in removing a square-root from the denominator, which is the aim of this video.
• Do you use the same pattern to rationalize a numerator?
• You rarely want to rationalise a numerator, but I've seen a couple of circumstances in which it is useful, and indeed these same techniques apply.
(1 vote)
• Why can (1-(Sqr root)2)/-1 equal out to (sqr root) 2 -1?
• both of them are the same, just multipled by -1.

(1 - √2)/-1 = -1 + √2, is the same propety as 1/-1 = -1/1
What is the propety?
Every positive multipled by a negative number is a negative number
And every negative number multipled by a negative number is positive.
As we know dividing something is the same as multiplying by reciprocal of it, so the rule works for diving too.
• how do you do questions that has whole number times radical divided by whole number times radical. E.x 15(sqrt3) / 3(srqt8)
• Just rationalize the denominator, don't worry about the numerator. Thus:
15√3 ÷ (3√8)
First, simplify the radicals. √8 = (√4)(√2) = 2√2. Thus,
15√3 ÷ (3√8) = 15√3 ÷ (6√2)
Now simplify the integers (15/6) = (5/2). Thus,
= 5√3 ÷ (2√2)
Multiply numerator and denominator by √2
= 5(√3)(√2) ÷ (2√2)(√2)
= 5√6 ÷ (2*2)
= (5√6) / 4
• How do you subtract square roots in a denominator
• Kathie,
You can eliminate a square root from the denominator like this:
1 / √2
You multiply by √2 / √2
1 / √2 * √2 / √2Multiply the numerators together and denominators together.
(1*√2) / (√2*√2) Do the multiplication
√2 / 2

3√2/4 - 1/√2
you multiply the second fraction by √2/√2 to get
(3√2)/4 - √2/2 Now find a common denominator
(3√2)/4 - (2√2)/4 Now you have a common denominator so you get
(3√2 - 2√2)/4 which is
1√2/4 or just
√2/4

I hope that is of help to you.