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Are you certain your quadratic formula solutions are right? It looks like you ignored the 1/x part, and solved it for simple x? I'd expect a substitution like y=1/x, and then putting your found solutions there in one more step to find actual values of x...
Yes, I am 💯
😅😅😅@@superacademy247
x=-1/4,simple method in mind
Yes, correct
we can also multiply both sides by x^3 and find one solution of
80 x^3 - x + 1 = 0
by inspection
x=1/n where n is divisor of 80
x=-1/4
s=sum of roots=0
p=produst of roots = -1/80
x1= -1/4
x2=a+b
x3=a-b
2a=1/4 => a=1/8
x1*x2*x3= -(1/4)*(a+b)*(a-b)=-(1/4)*(a^2-b^2)=-1/80
a = 1/8, b = 1/8 i sqrt(11/5)
a = 1/8, b = -1/8 i sqrt(11/5)
Nice one 🥰
At time 9:32 in your video, you stated the quadratic formula solution as m = (-b +- ROOT(b^2 - 4ac)) / 2c. Surely the denominator should be 2a, NOT 2c? Am I missing something?
It's a special type of quadratic formula. Check out on this video: ua-cam.com/video/mDaXTf60ivo/v-deo.htmlsi=nWU_d7sJQDtYrkHZ
Thank you for your explanation; maybe it would have been good to explain that at the time - I suspect that the vast majority of people watching the video would have been totally ignorant of that formula.
The original formula still works but leads to a complex number as a denominator . . . meaning an extra step to rationalize
Why not let y=1/x? It would simplify the computation to find the roots of y then x.
80=16+64=(-4)^2-(-4)^3=1/x^2-1/x^3
1/x^2=(-4)^2,1/x=-4,x=-1/4
(1/x^2) ➖ (1/x^3)={ 1/x^2 ➖ 1/x^3}=x^1 (x ➖ 1x+1). 2^40 2^2^20 1^1^2^10 2^2^5 1^2^1 2^1 (x ➖ 2x+1).
why use quadratic with c in denominator instead of a?
I applied a special type of quadratic formula: ua-cam.com/video/mDaXTf60ivo/v-deo.htmlsi=nWU_d7sJQDtYrkHZ
x = -1/y ; 1/x = -y
(-y)² - (-y)³ = 80
y² + y³ = 8.(2 . 5)
y².(y + 1) = 4² . (4 + 1)
y=4 ; x = -1/4
(y³ + y² - 80) / (y - 4) = y² + 5y + 20
résolution complexe de y² + 5y + 20 = 0, puis:
x' = -conj(y')/|y'|² et x" = -conj(y")/|y"|²
Super Academy, as you're addressing an english-speaking audience, please pronounce 'b' as "bee", NOT "pee" (which you always seem to do). Perhaps you have a speech impediment which does not allow you to say "bee"? Can you discriminate between "bee" and "pee"?
a² - a³ = 80
a³ - a² + 80 = 0
a³ - (5a² - 4a²) + 80 = 0
a³ - 5a² + 4a² + 80 = 0
a³ - 5a² + 4a² + 80 + (20a - 20a) = 0
a³ - 5a² + 4a² + 80 + 20a - 20a = 0
a³ - 5a² + 20a + 4a² - 20a + 80 = 0
(a³ - 5a² + 20a) + (4a² - 20a + 80) = 0
a.(a² - 5a + 20) + 4.(a² - 5a + 20) = 0
(a + 4).(a² - 5a + 20) = 0
First case: (a + 4) = 0
a + 4 = 0
a = - 4 → recall: a = 1/x
→ x = - 1/4
Second case: (a² - 5a + 20) = 0
a² - 5a + 20 = 0
Δ = (- 5)² - (4 * 20) = 25 - 80 = - 55 = 55i²
a = (5 ± i√55)/2 → recall: a = 1/x
x = 2/(5 ± i√55)
First possibility: x = 2/(5 + i√55)
x = 2.(5 - i√55)/[(5 + i√55).(5 - i√55)]
x = 2.(5 - i√55)/[25 + 55]
x = 2.(5 - i√55)/80
→ x = (5 - i√55)/40
Second possibility: x = 2/(5 - i√55)
x = 2.(5 + i√55)/[(5 - i√55).(5 + i√55)]
x = 2.(5 + i√55)/[25 + 55]
x = 2.(5 + i√55)/80
→ x = (5 + i√55)/40
Great substitution 👌
@10:52 WRONG. Better luck next time. You failed.
let u=1/x , u^3 - u^2 +/- n*u + 80 = 0 , by fakt., multp. *4 ,
+1 +4 u^3+4u^2 -5u^2-20u +20u+80=0 ,
-5 -20 (u+4)(u^2-5u+20)=0 , u= -4 , 1/x=-4 , x= -1/4 ,
+20 +80 = 0 , u^2-5u+20=0 , u=(5 +/- i*V55)/2 , 1/x=(5 +/- i*V55)/2 ,
solu. , x= -1/4 , 2/(5+i*V55) , 2/(5-i*V55) ,
test --> x= -1/4 , (1/(-1/4))^2 - (1/(-1/4))^3=16-(-64) , 16+64=80 , same , OK ,
********* comment , 2/(5+i*V55)=(1/8)-(1/8)*i*V(11/5) , & 2/(5-i*V55)=(1/8)+(1/8)*i*V(11/5) , *********
comment , 2/(5+i*V55)=(1/8)-(1/8)*i*V(11/5) , & 2/(5-i*V55)=(1/8)+(1/8)*i*V(11/5) ,
Great 👌 solution
@@superacademy247 Thanks , they give very good tasks, I use factoring for integer roots, but I learn a lot from their quadratic transformations, it's very difficult...