11 is a prime number = 6+5 ( Difference should be one , for example 17 = 9+8 ,13 = 7+6 etc.) 9 x^4 - 60 x +36 -25 = 0 9 x^4 +36 = 60 x + 25 (3x^2+6)^2 = 36 x^2 + 60 x + 25 (3x^2 + 6)^2 = (6 x+5)^2 (3x^2+6 x+11) (3x^2 - 6 x +1) = 0 Now we can use quadratic formula to get roots. This method generally works for equation ax^4+bx+c = 0 provided 'a' is perfect square.
Rewrite the equation as (3x^2+p)^2 = 6px^2 + 60 x + p^2 -11, and require the right hand side to be a perfect square in x. This requires the discriminant to vanish: ∆ = 60^2 - 4*6p*(p^2-11) = 0, or p^3 - 11 p - 150 = 0 Since we suspect the original x-equation to be factorisable over the integers, we first check if p=6 is a solution to this p-equation. It is! Hence we can rewrite the original equation as (3x^2 + 6)^2 - (6x+5)^2 = (3x^2 + 6x+11) (3x-1)^2 = 0.
It's based on completing the square method in which we take half the coefficient of the middle term then we square it. Hope this explanation helps💪✅💕💯🤗🤩
11 is a prime number = 6+5
( Difference should be one , for example 17 = 9+8 ,13 = 7+6 etc.)
9 x^4 - 60 x +36 -25 = 0
9 x^4 +36 = 60 x + 25
(3x^2+6)^2 = 36 x^2 + 60 x + 25
(3x^2 + 6)^2 = (6 x+5)^2
(3x^2+6 x+11) (3x^2 - 6 x +1) = 0
Now we can use quadratic formula to get roots.
This method generally works for equation ax^4+bx+c = 0
provided 'a' is perfect square.
That's a great solution for this specific problem! 🤩💪This is a clever approach! Thanks for sharing your perspective. 🙏💯
Великолепное решение!!!❤
Excellent!
I'm glad you enjoyed the video! 😁💯💕😎🙏Thanks for the kind words! 😊
Rewrite the equation as
(3x^2+p)^2 = 6px^2 + 60 x + p^2 -11,
and require the right hand side to be a perfect square in x. This requires the discriminant to vanish:
∆ = 60^2 - 4*6p*(p^2-11) = 0, or
p^3 - 11 p - 150 = 0
Since we suspect the original x-equation to be factorisable over the integers, we first check if p=6 is a solution to this p-equation. It is! Hence we can rewrite the original equation as
(3x^2 + 6)^2 - (6x+5)^2 =
(3x^2 + 6x+11) (3x-1)^2 = 0.
Thanks for your shortcut 🙏👏💡😎💕💯
Let E= (3x^2)^2-60x+11=(3x^2+ax+p)(3x^2+bx+q);=> pq=11;=> p=11& q=1: => b=-a as x^3 =0; =>11bx+ax=-60x; =>a=6(b=-a);=> E=(3x^2+6x+11)(3x^2 -6x+1)
can every quartic with no x³ term be expressed as the product of matching quadratics
@davidseed2939
Yes. eg. For(x^2+ax+p)(x^2+bx+q) > x^3 term is ax(x^2)+bx(x^2)= (a+b)x^3 & if no x^3 term then (a+b)=0=>b= -a .
182.19
A can't understand the step on 4:28, explain me please!
It's based on completing the square method in which we take half the coefficient of the middle term then we square it. Hope this explanation helps💪✅💕💯🤗🤩
Horribly tortured algebra. Is this an admissions test or a torture chamber?☠️⚰️
I'll consider coming up with a nice method in a few steps . 🤣🤣🤣