Rationalizing the denominator involving more than one square root

In summary, to rationalize the denominator in these types of expressions, you can multiply the numerator and denominator by the conjugate of the denominator. For the first problem, this would be 2+7√2-5√3. For the second problem, you can use the fact that a^3+b^3=(a+b)(a^2-ab+b^2) to simplify the expression before rationalizing the denominator.
  • #1
SwimmingGoat
7
0
Here's my problems:
How might you "rationalize the denominator" if the expression is [tex]\frac{1}{2+7√2+5√3}[/tex] or [tex]\frac{1}{\sqrt[3]{2}+1}[/tex]?
I know that in typical problems where we rationalize the denominator, we simply have to multiply the denominator and numerator by the conjugate of the denominator. For example, if the denominator were 2+√5, I'd simply multiply the top and bottom by 2-√5.
I haven't been in any math classes for two years, and I'm taking my final math course, and I simply don't remember how to do things like this! I'm not looking for someone to solve the whole thing, but I need suggestions or ideas on how I can solve these problems... Thank you so much for your help in advance!
 
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  • #2
You can solve the first problem in two steps - rationalize away one root, simplify, then get rid of the other root.

The second problem needs a clever choice for an expansion.
What you need as result is something like ##\sqrt[3]{2}^3+1## (=3), and this can be written as ##\sqrt[3]{2}^3+1 = (\sqrt[3]{2}+1) \cdot x## where you have to find x. Expand the fraction with x, and you get a nice denominator.
 
  • #3
Thank you so much for your help! For the second one, I applied your idea with the fact that [itex]a^3+b^3=(a+b)(a^2-ab-b^2)[/itex]. I let [itex]a=1[/itex]and [itex]b=\sqrt[3]{2}[/itex].
 
  • #4
SwimmingGoat said:
Thank you so much for your help! For the second one, I applied your idea with the fact that [itex]a^3+b^3=(a+b)(a^2-ab-b^2)[/itex]. I let [itex]a=1[/itex]and [itex]b=\sqrt[3]{2}[/itex].

Correctly: [itex]a^3+b^3=(a+b)(a^2-ab+b^2)[/itex]

ehild
 
  • #5
Whoops! Thanks for catching that.
 
  • #6
SwimmingGoat said:
Here's my problems:
How might you "rationalize the denominator" if the expression is [tex]\frac{1}{2+7√2+5√3}[/tex]?
Do what you usually do. First multiply numerator and denominator by [itex]2+7√2-5√3[/itex]. This will get rid of the square root of 3 in the denominator, and leave only a term with the square root of 2 in the denominator. Then multiply numerator and denominator by the conjugate of that.
 

What does it mean to rationalize the denominator involving more than one square root?

When the denominator of a fraction contains multiple square roots, it is considered to be irrational. Rationalizing the denominator involves manipulating the fraction to remove these square roots and make the denominator a rational number (a number that can be expressed as a ratio of two integers).

Why is it important to rationalize the denominator involving more than one square root?

Rationalizing the denominator can make it easier to work with fractions and to perform operations such as addition, subtraction, multiplication, and division. It also allows us to write fractions in their simplest form.

How do you rationalize the denominator involving more than one square root?

To rationalize the denominator, we multiply both the numerator and denominator by a suitable expression that will eliminate the square roots in the denominator. This can be done by multiplying by the conjugate (the same expression with the opposite sign between the two terms).

Can you provide an example of rationalizing the denominator involving more than one square root?

Sure! Let's take the fraction 1 / (√2 + √3). To rationalize the denominator, we multiply the fraction by (√2 - √3) / (√2 - √3). This gives us (1 * √2 - 1 * √3) / (√2 * √2 - √2 * √3 + √2 * √3 - √3 * √3), which simplifies to (√2 - √3) / (2 - 3), or (√2 - √3) / -1, which is equivalent to -√2 + √3. This is now a rationalized fraction with a rational denominator.

Is rationalizing the denominator involving more than one square root always necessary?

No, it is not always necessary. In some cases, leaving the denominator irrational may be more useful, such as when the square roots are needed for further calculations. However, in most cases, rationalizing the denominator is preferred.

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