# Tag Info

## Hot answers tagged radicals

9

Let $x=\sqrt2$ then we have $${x^4-1\over x^2+x+3}={x^6-x^2\over x(x^3+x^2+3x)}={4-x^2\over x(2+x^2+3x)}= {(2-x)(2+x)\over x(x+2)(x+1) }$$ $$= {2-x\over x^2+x}= {(2-x)(x-1)\over x(x+1)(x-1)}= {(2-x)(x-1)\over x^3-x}$$ $$= {(2-x)(x-1)\over 2-x} = x-1$$ Edit: but other solution is much nicer then this one.

8

From $6 = 2\times 3$, $10 = 2\times 5$, $15 = 3 \times 5$, observe that $10 + 2(\sqrt6 + 2\sqrt{10} + 2\sqrt{15}) = (\sqrt2+\sqrt3+\sqrt5)^2$.

6

With the same notation as in the other answer, i.e. $x = \sqrt 2$, noting that $x^3+1=3,$ you can write your quantity as \begin{eqnarray} \frac{x^4-1}{x^2+x+3} &=& \frac{(x-1)(x^3+x^2+x+1)}{x^2+x+x^3+1}=\\ &=&x-1. \end{eqnarray}

5

The question can be rephrased in abstract form as: If we have an equation of the form $$a^2=b^2$$ why is it equivalent to $a=\pm b$? Why not $\pm a = \pm b$? As Dr. Sonnhard Graubner's answer outlined, it can be explained by \begin{align*} &a^2=b^2\\[4pt] \iff\;&a^2-b^2=0\\[4pt] \iff\;&(a-b)(a+b)=0\\[4pt] \iff\;&a-b=0\;\;\;\text{or}\;\;\;a+...

5

The only incorrect statement is that $$\sqrt{e^{i2\pi}}=e^{i\pi}$$ Because we have that $$\sqrt{x^2}=|x|$$ for real $x$. So we would have the answer as $$\sqrt{e^{i2\pi}}=|e^{i\pi}|=|-1|=1$$

5

Hint: $\displaystyle\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}=\frac1{\sqrt n}\times\frac1{\sqrt{1+\frac1n}+1}.$

4

Hint. By the change of variable $$u=\frac12-x,\quad du=-dx,$$ observing that $\text{sign}\left(\frac{1}{2} - x\right)=1$ for $0<x<\frac12$, one just gets $$\int_0 ^{\frac{1}{2}} \text{sign}\left(\frac{1}{2} - x\right) \frac{1}{\left(\frac{1}{2}-x\right)^s} \text{d}x = \int_0 ^{\frac{1}{2}}\frac{du}{u^s},\quad s \in (0,1) .$$

4

Hint: $\displaystyle\lim_{x\to2}\frac{\sqrt{3x-5}-\sqrt{x-1}}{x-2}=\lim_{x\to2}\frac{\sqrt{3x-5}-1}{x-2}-\lim_{x\to2}\frac{\sqrt{x-1}-1}{x-2}.$

4

I will mention one (easily corrected) logical error and one stylistic piece of advice that could make the proof more readable. But the upshot is that this is a well-argued proof by any standard, and especially impressive for a first effort. When you said that $A^2$ and $B^2$ share no factors aside from 1, that does not imply that $\frac{A^2}{B^2}$ is not ...

4

Since $4-9/2 <0$, it's not true that $4-9/2=\sqrt{(4-9/2)^2}$

3

Set $y-3=z,$ then square both sides and simplify to get $$z^2-3z-8=-\sqrt{4+(z-3)^2}\sqrt{16+z^2}.$$ Squaring again and simplifying the right hand side gives $$(z^2-3z-8)^2=(z^2-6z+13)(16+z^2).$$ Now let $16+z^2=w,$ then expand and simplify to get $$z^2+16z+64=5w.$$ Substituting back for $w,$ simplifying and factoring gives the quadratic $$(z-2)^2=0.$$ Thus, ...

3

Your expression can be verified by the following identity: \begin{align} & \; \sqrt{a}+\sqrt{b}+\sqrt{c} \\ = & \: \dfrac {(a+b+c)^3-27abc} {\left( \sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}- \sqrt{bc}-\sqrt{ca}-\sqrt{ab} \, \right) \left[ (a+b+c)^2+ 3(a+b+c)\sqrt{abc}+ 9\...

3

Notice that $$a-b = (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})$$ so you get $$(\sqrt{a}-\sqrt{b}+\sqrt{b})/(\sqrt{a}-\sqrt{b}-\sqrt{a})=-\sqrt{a\over b}$$

3

There is nothing wrong with the equality $1=\sqrt{(-1)(-1)}$. The problem lies in the equality $\sqrt{e^{2\pi i}}=e^{\pi i}$, because $e^{2\pi i}=1$, and therefore $\sqrt{e^{2\pi i}}=1\neq e^{\pi i}$.

3

One thing right off the bat. Every non-zero complex number will have two square roots so there wont be just $x + yi$ there will also be $-x -yi$ " using geuss and check" Why use guess and check when you can use the quadratic formula? $a^2 - b^2 = 10$ and $2abi = -4\sqrt 6i$ $a=\frac{-2\sqrt 6}b$ $\frac {4*6}{b^2} - b^2 = 10$ $24 - b^4 = 10b^2$ $b^4 + ... 3 Let $$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2=x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}$$ So $$\sqrt{x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}}=\sqrt{x}+\sqrt{y}+\sqrt{z}$$For $$\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{x}+\sqrt{y}+\sqrt{z}.$$We get$x+y+z=10, 2\sqrt{xy}=\sqrt{24}, 2\sqrt{yz}=\sqrt{40}, 2 \sqrt{xz}=\sqrt{60}.$So we get$xy=6, yz=10, zx=15 \Rightarrow ...

3

The fallacy is in thinking $\sqrt{x^2}=x$. Actually, $\sqrt{x^2}=|x|$.

2

We have that $$a^2-b^2=10 \quad \textrm{and} \quad 2ab=-4\sqrt 6$$ Now, squaring both equalities and addem up we get $$(a^2+b^2)^2=(a^2-b^2)^2 +(2ab)^2=10^2+(-4\sqrt 6)^2 =196$$ $$\Rightarrow \quad a^2+b^2=14$$ and using again the first equality we obtain $$a^2=12 \quad \textrm{and} \quad b^2=2$$ or $$a=\pm 2\sqrt 3 \quad \textrm{and} \quad b=\pm \sqrt 2$$ ...

2

If $\sqrt{n-1}+\sqrt{n+1}$ is rational and $n>1$ then $(\sqrt{n-1}+\sqrt{n+1})^2=2n+2\sqrt{n^2-1}$ is rational. Then $\sqrt{n^2-1}$ is rational. Then $n^2-1=\frac{p^2}{q^2}$, where $p,q \in \mathbb{N}$, $gcd(p,q)=1$. Then $q=1$. Then $n^2-1=p^2$. But $n^2-1=p^2$ hasn't solutions.

2

The sum of two irrational numbers can be rational (e.g. $x,\,q-x$ with $q\in\Bbb Q\not\owns x$). Their strategy was to note that, since $\sqrt{n-1}\pm\sqrt{n+1}\in\Bbb Q$, taking linear combinations thereof gives $\sqrt{n\pm 1}\in\Bbb Q$.

2

One thing we can do is render the third superroot as an iteration of second superroots. Let $x^{x^x}=a$. Then $(x^x)^{(x^x)}=a^x$ and we take two square superroots to get a fixed point iteration: $\color{blue}{x=\sqrt{\sqrt{a^x}_s}_s}$ If we put $a>1$ and $x=1$ on the right side, we get $x=\sqrt{\sqrt{a}_s}_s$ as our next iteration, and this will ...

2

Consider points $A(2,6)$, $B(-4,3)$ and a point $P(0,y)$ on y-axis. Then notice that given equation is actually $$AP+BP= AB$$ so $P$ is (by triangle inequality ) on a line $AB$ so it is an intersection point of the line $AB$: $$y= {1\over 2}x+5$$ and $y$-axis, so $y=5$.

2

If $n$ is odd, $x^n$ is an invertible function from $\Bbb R$ to itself; we denote the inverse either $\sqrt[n]{x}$ or $x^{1/n}$. This is consistent with $\left(x^a\right)^b=x^{ab}$. We can now uniquely define $x^{p/q}\in\Bbb R$ for any $x\in\Bbb R$ with $x\ne0$ (a restriction we can drop if $p/q\gt0$), and any integers $p,\,q$ with odd $q>0$. It won't ...

2

This is a redo of a previously posted answer. This one attempts to be more approachable to readers without knowledge of finite field extensions. The $\sqrt{16}$ and $\sqrt{27}$ are distractions. Write these as $2\sqrt{2}$ and $3$. So you have $$\frac{-1+2\sqrt{2}}{2+\sqrt{2}+\sqrt{4}}$$ All five terms are in the form $a\sqrt{2^n}$. (For ...

2

You made a mistake in your derivation. You should have $(x^2+y^2)^\color{red}2=a^2+b^2$. This follows from $a=x^2-y^2$ and $b=2xy$ (or from known properties of complex modulus). Thus, $a=x^2-\dfrac {b^2}{4x^2}$; solving this quadratic equation in $x^2$ yields $x^2=\dfrac{a+\sqrt{a^2+b^2}}2$ as the correct answer.

1

Let $y=x-2$, then $$L=.\lim_{y\rightarrow 0} \frac{(1+3y)^{1/2}-(1+y)^{1/3}}{y}$$ Use binomial approximation $(1+z)^p \approx 1+pz$ if $|z|<<1$. Then $$L=\lim_{y \rightarrow 0} \frac{(1+3y/2)-(1+y/3)}{y}= \lim_{y\rightarrow 0} \frac{7y}{6y}=7/6.$$ Since it is $0/0$ form you may also use L'Hospital rule, differentiate up and down separately: $$L=\... 1 If there exists a cubic polynomial of integer coefficients with roots:$$0,\sqrt {A-15√3} , \sqrt {A+15√3} $$then it will be of the form x^3-4x^2 + \sqrt{A^2-675}x You just need to find A such that A^2-675 is a perfect cube. Trivially A^2 = 676 will do. 1 Let x=A-15\sqrt 3 and y=A+15\sqrt 3, then the equation becomes$$x^{1/3}+y^{1/3}=4.$$Taking cubes of both sides gives$$x+y+3(xy)^{1/3}(x^{1/3}+y^{1/3})=4^3.$$Now since x+y=2A, and x^{1/3}+y^{1/3}=4, the equation becomes$$2A+3(xy)^{1/3}(4)=4^3,$$which gives$$6(xy)^{1/3}=32-A.$$Now, cubing and substituting for xy=A^2-15^2\cdot 3 gives$$6^3(A^...

1

Let $x,y$ be the numbers $(A\pm 15\sqrt 3)^{1/3}$ so that $x+y=4$. hen we have: \begin{aligned} 2A &= x^3+y^3\\ &= (x+y)^3-3xy(x+y) \\ &= 4^3-3xy\cdot 4 \\ &=64 - 3(A^2-675)^{1/3}\cdot 4\ .\text{ So:} \\[3mm] 6^3(A^2-675) &=(32-A)^3\ . \end{aligned} This gives us an equation in $A$, that we may solve (with bare hands or not). sage: ...

1

\lim_{n \to \infty}\sqrt{n} \cdot (\sqrt{n+1} - \sqrt{n})= \lim_{n \to \infty}\sqrt{n} \cdot \frac{(\sqrt{n+1} - \sqrt{n})}{1}= \lim_{n \to \infty}\sqrt{n} \cdot \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{(\sqrt{n+1} + \sqrt{n})}= \lim_{n \to \infty}\sqrt{n}\cdot \frac{n+1-n}{\sqrt{n+1}+\sqrt n}= \lim_{n \to \infty}\sqrt{n}\cdot \frac1{\sqrt{n+...

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