How Does 2/sqrt 2 = Sqrt 2

Why does -2 + sqrt (6+4*sqrt(2)) = sqrt(2), or more so, how does it I can't see why it works?

Let’s say you did not know that it was equal to [math]\sqrt{2}[/math], but instead you were looking to solve this problem:Simplify [math]-2 + \sqrt{6+4\sqrt{2}}[/math]How would you do that? First, notice the ugly part is the [math]\sqrt{6+4\sqrt{2}}[/math], so really this is a problem to simplify that.We can take the 4 back inside the radical to make this into: [math]\sqrt{6+\sqrt{32}}[/math]When you have the square root of 2 terms where one term is an integer and one is a square root, you can hypothesize that it is equal to the square root of 1 number plus the square root of another number.Therefore, I hypothesize [math]\sqrt{6+\sqrt{32}} = \sqrt{a} + \sqrt{b}[/math]Now we square both sides.[math]6 + \sqrt{32} = a + 2\sqrt{ab} + b[/math][math]6 + \sqrt{32} = a + \sqrt{4ab} + b[/math]On both sides of the equation we have a square root, and we have something that is not a square root. We could write those as a system of 2 equations.[math]\sqrt{32} = \sqrt{4ab}[/math][math]6 = a + b[/math]From the first equation, we have[math]32 = 4ab[/math][math]8 = ab[/math]From the second equation we have[math]a = 6 - b[/math]Substituting that into the first equation we get:[math]8 = (6-b)b[/math][math]b^2 - 6b + 8 = 0[/math]Using the handy dandy quadratic formula we get:[math]b = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)}[/math][math]b = \dfrac{6 \pm \sqrt{36 - 32}}{2} = 3 \pm 1[/math]Therefore we have 2 possible values of b, 2 or 4. If [math]b=2[/math], then:[math]a = 6 - 2 = 4[/math]If [math]b=4[/math], then:[math]a = 6 - 4 = 2[/math]Therefore, you can see those 2 solutions are equivalent, the only difference is the order of the terms. Recall that to get those values, we started with this hypothesis:[math]\sqrt{6+\sqrt{32}} = \sqrt{a} + \sqrt{b}[/math]Therefore, we now have:[math]\sqrt{6+\sqrt{32}} = \sqrt{4} + \sqrt{2}[/math]Knowing that, we can now simplify the original expression.[math]-2 + \sqrt{6+4\sqrt{2}}[/math][math]-2 + \sqrt{4} + \sqrt{2}[/math][math]\sqrt{2}[/math]Therefore, I have shown that [math]-2 + \sqrt{6+4\sqrt{2}} = \sqrt{2}[/math] without knowing what the expression simplified to beforehand.

How do I integrate arcsin(x/2) / (sqrt(2-x))?

Let [math]\displaystyle I = \int \dfrac{\arcsin(\frac{x}{2})}{\sqrt{2 - x}} \, dx[/math]Lets apply the integration by parts technique,Assume [math]u = \arcsin(\frac{x}{2})[/math][math]\implies du = \dfrac{1}{2\sqrt{1 - \frac{x^2}{4}}} \, dx[/math][math]\implies du = \dfrac{1}{\sqrt{4 - x^2}} \, dx[/math]and [math]dv = \dfrac{1}{\sqrt{2 - x}} \, dx[/math][math]\implies v = \int \dfrac{1}{\sqrt{2 - x}} \, dx[/math][math]\implies v = -2\sqrt{2 - x}[/math]As, [math]\int u \, dv = uv - \int v \, du[/math]So, [math]\displaystyle I = -2\sqrt{2 - x}\arcsin(\frac{x}{2}) + \int \dfrac{2\sqrt{2 - x}}{\sqrt{4 - x^2}} \, dx[/math][math]\displaystyle = -2\sqrt{2 - x}\arcsin(\frac{x}{2}) + \int \dfrac{2}{\sqrt{2 + x}} \, dx[/math][math]\displaystyle = -2\sqrt{2 - x}\arcsin(\frac{x}{2}) + 4\sqrt{2 + x} + C[/math] (where [math]C[/math] the arbitrary constant of indefinite integration)

Why is [math]2\sqrt 2=\sqrt8[/math]?

Hard to explain here but I'll try…2√2 is just a simplified way of writing √8, like 12 is just a simplified way of writing √144Why?√8 is a surd. This means it is an irrational number (one that effectively does not end , like 1/3 is 0.3333333333333… indefinitely)We can simplify surds, as shown above.First we must clairfy this: like 8 = 4 x 2, √8 = √4 x √2.This can be written as √4√2Remember we are simplifying. Is there any way to simplify further √4√2?Yes - √4 is simplified to 2.Consequently, √4√2 = 2√2How to we choose which numbers to simplify the surd to?Let's take the surd √200The rule is: factorise the surd by taking its largest square number factors.List of factors of 200 - 2, 4, 5, 10, 20, 25, 40, 50, 100 (there are others but none greater than two hundred)100 is the largest square factor200 = 100 x 2 …√200 = √100 x √2√200 = √100√2SIMPLIFY√100 = 10… so√200 = 10√2Hope this helps!

What is 2^sqrt2?

let 2^(√2) be xlog x = log 2^(√2)log (a^b) = b * log atherefore log x = √2 log 2now we need a log table or a calculator to calculate the valueslog x = 1.414 * 0.301log x = 0.4256x= antilog 0.4256 = 2.664