Great work! As a college lecturer of mathematics (and also engineer), I always avoid even writing compound numbers, like the 7 1/2 given partway through this video -- compound numbers are ambiguous. Compound numbers should have no place in higher mathematics. In cooking recipes, fine, but not higher math. Also, I always teach the formula that "if a^2 = b, then a = +/- sqrt(b)", where the +/- is applied immediately. This was skipped for one step.
Solving the equation will be greatly simplified after the substitution y= x-9 (y-1)²+y³+(y+1)⁴=2 y²-2y+1+y³+y⁴+4y³+6y²+4y+1=2 y⁴+5y³+6y²+2y=0 The first root is y=0, x=9 y³+5y²+6y+2=0 The next root will be searched from integer dividers of the last coefficient 2, i.e. y=±1,±2, checking only negative ones y=-1,-2 So y=-1 is the next root (x=8) Dividing the polynomial by (y+1) we'll get the quadratic equation y²+4y+2=0 (y+2)²=2 y=-2±√2, x=7±√2 So we've got 4 roots: y=0,-1,-2±√2 x=9,8,7±√2
@@LooneyKids847 sqrt(x) is always non-negative in IR, that's why we write a = +/- sqrt(1), but that's long form of a = -1 v a = 1. This is obviously a polynomial of grade 4, that has exactly 4 solutions (and, see, all of them are real, even positive!)...
Great work!
As a college lecturer of mathematics (and also engineer), I always avoid even writing compound numbers, like the 7 1/2 given partway through this video -- compound numbers are ambiguous. Compound numbers should have no place in higher mathematics. In cooking recipes, fine, but not higher math. Also, I always teach the formula that "if a^2 = b, then a = +/- sqrt(b)", where the +/- is applied immediately. This was skipped for one step.
Great tip!
Thank you!
Solving the equation will be greatly simplified after the substitution y= x-9
(y-1)²+y³+(y+1)⁴=2
y²-2y+1+y³+y⁴+4y³+6y²+4y+1=2
y⁴+5y³+6y²+2y=0
The first root is y=0, x=9
y³+5y²+6y+2=0
The next root will be searched from integer dividers of the last coefficient 2, i.e.
y=±1,±2, checking only negative ones
y=-1,-2
So y=-1 is the next root (x=8)
Dividing the polynomial by (y+1) we'll get the quadratic equation
y²+4y+2=0
(y+2)²=2
y=-2±√2, x=7±√2
So we've got 4 roots:
y=0,-1,-2±√2
x=9,8,7±√2
I would have taken a = x - 9. My intuition is, that it will solve nicely... BTW: First obvious solution is x = 9...
A substituição de variáveis é uma forma elegante de resolver essa equação.
Cool
Thanks!!👍👍
Bravo.
Thank you very much!
Don't do square root both side like that.
If a²= b, a1 = +√b, a2= -√b, because radical √b always positives.
He never takes care of this... But he can take 30s to explain why "x-8=y x=y+8)". Mysteries...
9 by inspection.
Yes
9
The one of root of this equation is 9
Use intuition
Bu kadar işleme ne gerek var
Sonuç iki ise Üç işlemden bir tanesi -1 bir tanesi 0 birtanesi ise +1 olacak üsler zaten çift bu sebebten x=9 çıkar
X=9
İşlem yapmadan kafamdan yaptım x=9
7:43 It's not ±1. It's only +1.
No, it's right.
Because we can solve (a² - 1) to (a + 1) * (a - 1). There are 1 and -1, so it's right
No because you can do -1 * -1 = 1
That is right but when we use roots we only want the positive answer.
yes but for finding zeroes we go both side of the num line@@LooneyKids847
@@LooneyKids847 sqrt(x) is always non-negative in IR, that's why we write a = +/- sqrt(1), but that's long form of a = -1 v a = 1.
This is obviously a polynomial of grade 4, that has exactly 4 solutions (and, see, all of them are real, even positive!)...
Х=9
x=9