No no don't think it is an assumption people make It is a written rule, the sqrt function is defined that way sqrt(x^2) = |x| Hence sqrt(3^2) = 3 sqrt((-3)^2) = 3
@@zhunya2495 You could be correct but I am not well versed in complex domain. I have only dealt and learnt about the behaviour of things over R by myself
Since sqrt(x^2)=x^((2)(1/2))=x^((1/2)(2))=(sqrt(x))^2, which is always positive for real numbers. So for real numbers, sqrt(x^2) = |x|, but for imaginary numbers the square can be negative so the absolute value does not apply. In general, when dealing with square roots of positive real numbers, the convention is to use the positive root.
The first one is straightforward: sqrt(2x+5) is sqrt(-1), so 2x=-6 which means x=-3. Now for the second: sqrt(2x+5)=-1. At first glance, squaring both sides makes sense (i.e. 2x+5=1 therefore 2x=-4, so x=-2). The problem is that solution of the square root function is always a positive root...Maybe if we change the expression on the right to i^2?. P.S., this is my thoughts as of not having watched the video yet..
I think the second answer comes from the fact that i⁴=1, take 1/4 power on both side, ⁴√1=i, replace with double square root, √√1=i, and square both sides, you get √1=-1. Effectively it comes from the fact that -1 is one of the 4th roots of 1. Cool
Sory gays you are not correct on the second one becouse the two equations are ligt the two answet are correct becouse the square Root of 1 are the absolute values of 1 mens -1or+1
Why is the square root of a positive number positive by definition? This is purely a matter of convention. There is no deep mathematical principle being revealed by saying √(1) = 1 but √(1) ≠ -1. The whole issue looks to me like the squabble about whether 6÷3(2) is 1 or 4. It's a fundamentally stupid question, revealing nothing of any worth, just a picky definition. (Not to say the video was stupid. It wasn't.)
@@jairogen90 Yo no intenté de resolver la ecuación, Wolfram también afirma que no tiene solución. Apenas sustituí la x por ese valor en la ecuación, pero no me dio -1, me dio (-i+1)/2^½.
@@josefonsecadecarvalhoolive1992 Sustituyendo mi resultado en la ecuación original da -1. También he probado otro método con resultado satisfactorio. Primero sustituyo -1 por i^2 y después elevo ambos miembros al cuadrado, quedando 2X+5=i^4. Despejando queda X=(i^4-5)/2. Creo que esta bien, corregidme si me equivoco.
@@paradiseexpress3639 I'm saying that everyone can watch this. From beginners to experts. From students to teachers. And he didn't mention anything about e^iπ
