Linked Questions

1 vote
5 answers
489 views

Simplify $\sqrt{8-\sqrt{63}}$ [duplicate]

I simplified the expression into $$\sqrt{8-3\cdot \sqrt{7}}$$ but my tutor said it wasn't the answer he was looking for. Can someone help me?
Arthur's user avatar
  • 51
0 votes
3 answers
243 views

Method for denesting square roots [duplicate]

Does anyone have a quick method of denesting square roots? Problem: To quickly denest $\sqrt{5+2 \sqrt{6}} \qquad \tag{1}$ As an aside, this comes from solving the polynomial $$x^4-10x^2+1=0$$ ...
user avatar
-1 votes
4 answers
158 views

How to transform $\sqrt{5-2\sqrt{6}}$ to $\sqrt{3}-\sqrt{2}$ by arithmetic operations? [duplicate]

Three sides of a right-angled triangle are $2\sqrt{3}$ (hypotenuse), $\sqrt{6}+1$, & $\sqrt{5-2\sqrt{6}}$. How do I transform the third side's length to $\sqrt{3}-\sqrt{2}$ for making my life ...
tryingtobeastoic's user avatar
2 votes
2 answers
152 views

What is an algorithm to test whether a number field element has a square root in that number field? [duplicate]

Let $K$ be a number field, such as $\mathbb{Q}(\sqrt{d})$ for $d$ a square-free integer. I am looking for an algorithm that outputs whether some $\alpha \in K$ has a square root in $K$, i.e. when ...
Sam Freedman's user avatar
  • 3,999
2 votes
2 answers
132 views

How can I prove algebraically that $\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$? [duplicate]

I found out that $\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$, by reverse engineering: $$\sqrt{6}-\sqrt{2} = \sqrt{ (\sqrt{6}-\sqrt{2} )^{2} } = \sqrt{ (\sqrt{6}-\sqrt{2} ) (\sqrt{6}-\sqrt{2} )}= \sqrt{ 6-...
Boas's user avatar
  • 29
0 votes
2 answers
97 views

How to take the square root of $BC$ where $BC^2$ equals $56-32\sqrt3$ [duplicate]

$BC^2$ equals $56-32\sqrt3$ what is the square root of BC? The dimensions I used to get this far (and I confirmed that they are correct) are A,B,C, in a right triangle, where B is $\sqrt{(42-24\sqrt3)}...
TizzleRizzle's user avatar
1 vote
0 answers
78 views

Is there any general way of finding root of a irrational number? (eg, getting $\sqrt2+\sqrt3$ from $\sqrt{5+2\sqrt6}$) [duplicate]

for example: $(\sqrt2+\sqrt3)^2 = 5+2\sqrt6$ is straightforward. But how can we get $\sqrt2+\sqrt3$ from $\sqrt{5+2\sqrt6}$? Is there any general method to do it?
ICFSZ's user avatar
  • 31
2 votes
0 answers
74 views

When is $\sqrt{n-\sqrt{m}}=\sqrt{a}-\sqrt{b}$ where $m,n\in\Bbb{Z}^+$ and $a,b\in\Bbb{Q}$? [duplicate]

In this youtube video, Pratik Matematik asked "$\sqrt{\sqrt{49}-\sqrt{48}}=?$". I generalized it as $\sqrt{n-\sqrt{n^2-k^2}}=\sqrt{\frac{n+k}{2}}-\sqrt{\frac{n-k}{2}}$. And the solution of ...
Bob Dobbs's user avatar
  • 11.9k
0 votes
0 answers
51 views

Is there formula to easily factorize $7+4 \sqrt{3}$ to $(2+ \sqrt{3} )^2$? [duplicate]

Is there some formula to easily guess that: $7+4\sqrt{3}$ can be factorize to: $(2+ \sqrt{3}) ^2$? Only looking at $7+4\sqrt{3}$, it's difficult to guess that you can factorize like that. EDIT: ...
ThePhi's user avatar
  • 170
0 votes
0 answers
53 views

Simplifying the square root [duplicate]

The question is "Find the exact value of $\sqrt{97+56\sqrt{3}}$ ". It's from some regional contest back in 2013 and the answer is $7+4\sqrt{3}$. Can someone explain how they can reach the ...
Algi King's user avatar
0 votes
0 answers
43 views

Question about exponent laws [duplicate]

I can't seem to demonstrate how $\frac{\sqrt{\sqrt{3}+2}}{2}$ is equal to $\frac{\sqrt{2}+\sqrt{6}}{4}$... even though my calculator tells me that both are equal $0.965925826$. Hoping someone can show ...
sesandc3123's user avatar
27 votes
7 answers
12k views

Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$

Here is the question: $$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$$ (original image) I think ...
frukoprof's user avatar
  • 413
16 votes
6 answers
15k views

Simplifying $\sqrt[4]{161-72 \sqrt{5}}$

$$\sqrt[4]{161-72 \sqrt{5}}$$ I tried to solve this as follows: the resultant will be in the form of $a+b\sqrt{5}$ since 5 is a prime and has no other factors other than 1 and itself. Taking this ...
1110101001's user avatar
  • 4,198
89 votes
3 answers
5k views

Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$

The following result discussed by Ramanujan is very famous: $$\sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{\frac{1}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}\tag {1}$$ and can be easily proved by ...
Paramanand Singh's user avatar
  • 88.3k

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