I Solved A Challenging Radical System in Two Ways

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11 дек 2023

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Комментарии : 18

@turbavykas 7 месяцев назад

Don't we loose solutions when we use sqrt(x)=a? Because x can be negative if z and y are negative. In our case equation 1 and 3 not possible but general you have to be careful.

@luunguyen193 Месяц назад

In Vietnam, we're learnt that we should do this thing fist! x-sqrt(yz)=1sqrt(yz)=x-1 yz=(x-1)^2 (with conditions yz>=0 and x-1>=0) The same as the others and we can continue to solve it :)

@xsimox13 7 месяцев назад

You can get the linear relationship much quicker by multiplying the first equation by a , the second by b, and the third by c and you add them all up together you get directly: a^3+b^3+c^3 - 3 abc = a - 5b + 7c

@dmtri1974 7 месяцев назад

If x>0 (then y, z >= 0) the system is congruent to a linear system like this : From the given equations we get x -1=sqrt(yz) (1) y +5=sqrt(xz) (2) z -7=sqrt(xy) (3) . Multipliyng (1) and (2) we get (x-1) (y+5) = z sqrt(xy) = (sqrt(xy)+7) sqrt(xy) => xy+5x-y-5 = xy +7sqrt(xy) => 5x-y-5 = 7(z-7) => 5x-y-7z =-44 (4) . By analogous treatment from (1) and (3) and (2) and (3) we also get the equations -7x+5y-z=-32 (5) -x-7y+5z=34 (6) . The linear system of (4), (5), (6) gives the solution x=4, y=1, z=9 . Furthermore, if x< = 0 then from the 1st we get no solution.

@anonymouschessaccount5508 7 месяцев назад

Your content is awesome 😊 and you deserve to get more popularity. With those aside, there's a slight mistake in the video. 8:04 you said that if *√x = -2* then there won't be real solutions for x, just complex ones. But, there won't be complex solutions either, and that equation has ABSOLUTELY NO SOLUTION. If we are to solve that, √x = -2 => x =⁰ (-2)² => x =⁰ 4 Plugging the values in, We get √4 =⁰ -2. This is not real as the square root of a number which is not in equation, has the main root only as their answer. If you go to the complex world, there would be no number that can fit √x = -2 The general rule is, If x² = -(|a|), then x = (√a)i ...[i² = -1] If √x = -(|a|), then x doesn't have ANY solution. [ *| |* represents the absolute value of a number ] Yeah that's it, the video is awesome 👌

@SyberMath 7 месяцев назад

I agree! Thanks

@adamsmithson486 7 месяцев назад

Pozdrawiam serdecznie i życzę miłego dnia

@vcvartak7111 7 месяцев назад

Not understood your second method what is to be done with (a^3+b^3+c^3 - 3abc) is it to factories?

@SyberMath 7 месяцев назад

No since all variables will contain the same expression, you can call that k and write each variable as a multiple of k and then substitute that into one of the equations