# Tag Info

### Is enumerating squares via $(a+1)^2 = a^2+(2a+1)$ a new method to find square roots?

This is not a new method. It's essentially: guess if the guess was too small, guess one bigger (or vice versa) repeat This is just a bit better than blind trial-and-error, and would have been ...

Accepted

Accepted

1 vote

### If $\alpha^3-\alpha+1=0$ then $\sqrt{3}\neq a\alpha^2+b\alpha+c$ for $a$, $b$, $c\in \mathbf{Q}$

Just to finalise what's been commented. Checking irreducibility of $f$ is maybe easier than you think! $f$ is cubic, checking irreducibility amounts to checking no rational roots exist. Were $f$ ...
1 vote
Accepted

### Confused about the extraneous root of $\frac{x \sqrt{A^2 - x^2} + x}{x^2 - \sqrt{A^2 - x^2}}$

$\sqrt{A^2 - x^2} = -1$ has no real solution. LHS is defined only when $A^2 - x^2 \ge 0$ in which case, LHS is non-negative and so, the equation has no real solution. Thus, $x=0$ is the only real ...
1 vote

### Is enumerating squares via $(a+1)^2 = a^2+(2a+1)$ a new method to find square roots?

You got it wrong. She inspired her teacher to find this: Suppose we want to find out that 21904 = 148² We guess 99² and get 9801 We miss by |21904 - 9801| = 12103 Dividing by twice the guess we get ...
1 vote

### Is enumerating squares via $(a+1)^2 = a^2+(2a+1)$ a new method to find square roots?

It looks like a geometrical insight: the total surface equals the brown, plus two times one yellow plus the green: (the yellow ones are $1$ large). The fact that a child (11 years old) has seen this, ...
1 vote

### Solving $y+\sqrt{y^2-1}=e^x$ respect to $y$

I want to present a solution, albeit not as elegant as others; hopefully, you will feel you could have discovered it. You have already tried using some trigonometric function to substitute $y$. I ...
1 vote

### $f(x) = \sqrt x^{{\sqrt{x}}^{\sqrt{x},\cdots}}$ asymptotic?

With Mathematica there is a loglog plot \text{ListPlot}\left[\text{Table}\left[\left\{x,(\log (\text{$\#$1}+1\&)\left((\log (\text{$\#$1}+1)\&)\left(\text{Fold}\left[\text{$\#$1}^{x^{\frac{1}...

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