For questions on rationalising the denominator, the operation of rewriting a fraction in such a way that the denominator is free of square roots, cube roots, etc. The fraction can be a real number involving radicals, but also a function.

# Questions tagged [rationalising-denominator]

75 questions

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votes

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### Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$

We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this…

欲しい未来

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### Rationalizing the denominator having square roots and cube roots

In middle-school mathematics, the teachers always tell you that if you have radicals on the denominator of a fraction, then it isn't fit to be a final answer - you have to rationalize the denominator, or get rid of all of the radicals in the…

Franklin Pezzuti Dyer

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### Rationalization of $\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$

Question:
$$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$ equals:
My approach:
I tried to rationalize the denominator by multiplying it by $\frac{\sqrt{2}-\sqrt{3}-\sqrt{5}}{\sqrt{2}-\sqrt{3}-\sqrt{5}}$. And got the result to be (after a long…

Gaurang Tandon

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### How do I rationalize the following fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$?

As the title says I need to rationalize the fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$. I wrote the denominator as: $\sqrt[3]{9^4}-\sqrt[3]{9^2}-3^3$ but I do not know what to do after. Can you help me?

Ghost

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### How do I simplify $\frac{\sqrt{21}-5}{2} + \frac{2}{\sqrt{21} - 5}$?

How do I simplify the following equation?
$$\frac{\sqrt{21}-5}{2} + \frac{2}{\sqrt{21} - 5}$$
I have no idea where to start. If I multiply either fraction by its denominator I will still end up with a square root. I know the end result should be…

user1534664

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### Simplifying $f(\sqrt{7})$, where $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$

If $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$ ; then $f(\sqrt {7})=\; ?$
I tried solving this equation through many methods, I tried rationalizing, squaring, etc. But after each of them, the method became really lengthy and ugly.
I also…

user880107

**6**

votes

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### How is $-16/4i$ equal to $4i$?

I came across a problem: $-16/4i$. Every time I put it into a calculator, it comes out as $4i$, but when I try to solve it is $-4i$, because of the negative one in front of the $16$.

Joce

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### Simplification of $\frac{2\sqrt{21}-\sqrt{35}+5\sqrt{15}-16}{\sqrt7+2\sqrt5-\sqrt3}$

Simplify
$$\dfrac{2\sqrt{21}-\sqrt{35}+5\sqrt{15}-16}{\sqrt7+2\sqrt5-\sqrt3}$$
Final solution should have rational denominators.
Suppose the solution is $X$, I have tried to make up an equation for $X^2$
$$X^2 =…

Cyh1368

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### Rationalizing $\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}$ in two ways gives different answers

I have a doubt see when we rationalize denominator of expression $$\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}- \sqrt{1-\cos x}}$$
we get answer $$\frac{1+\sin x}{\cos x}$$ but when we rationalize numerator we get
$$\frac{\cos x}{1+\sin…

Neev Garg

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**1**answer

### How do I simplify $34\csc{\frac{2\pi}{17}}$?

I have $$ 34\csc\left(\dfrac{2\pi}{17}\right)$$ is equal to $$\dfrac{136}{\sqrt{8-\sqrt{15+\sqrt{17} + \sqrt{34 + 6\sqrt{17} - \sqrt{34-2\sqrt{17} } + 2\sqrt{ 578-34\sqrt{17}} - 16\sqrt{34-2\sqrt{17}} } }}}.$$
I want to rationalize it, but I am…

slowerthanstopped

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### How to rationalize multiple terms with fractional exponents

I'm trying to derive the derivative of $f(x) = x^{2/3}$ using the limit definition:
$$f'(x)=\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
$$=\lim_{h \to 0} \frac{(x+h)^{2/3} - x^{2/3}}{h}$$
I suspect I have to rationalize the numerator in order to…

Slecker

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### How to show $\frac{1}{2\sqrt{2} + \sqrt{3}} = \frac{2\sqrt{2} - \sqrt{3}}{5}$?

Show that:
$$ \dfrac{1}{2\sqrt2+\sqrt3}=\dfrac{2\sqrt2-\sqrt3}{5}$$
So I multiplied everything by $\sqrt3$
Then I got
$$\frac{\sqrt{3}}{2\sqrt{2}+3}$$
Then multiply it by $\sqrt2$ to obtain
$$\frac{\sqrt{2}\sqrt{3}}{2 \cdot 3+3}$$
Which is…

JohnFire

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### how to rationalize $\frac {x-8}{\sqrt[3]{x}-2}$

In order to resolve a limit, I need to rationalize $\frac {x-8}{\sqrt[3]{x}-2}$. I tried multiplying it by $\sqrt[3]{x^3}$ or $\sqrt[3]{x^2}$ but with no much success. It seems that I can't use "Difference of Squares" identity too. The limit in…

**3**

votes

**4**answers

### Simplify $\frac{1}{\sqrt{-1+\sqrt{2}}}$

In an exercise I got as solution $\frac{1}{\sqrt{-1+\sqrt{2}}}$, it now holds that $\frac{1}{\sqrt{-1+\sqrt{2}}} = \sqrt{1+\sqrt{2}}$. But I really don't see how you can manipulate the left hand side to become the right hand side (quite shameful as…

Filip M

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### Solving this limit $\lim_{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$.

The question is $\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$.
I hope you guys understand why I have written the numerator like that. So my progress is nothing but $1+\frac{\sqrt{x+6}-3}{x^2-9}$.
Now how do I rationalize the numerator?
It…

Archis Welankar

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