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.
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.
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💪✅💕💯🤗🤩
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 🙏👏💡😎💕💯
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! 😊
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💪✅💕💯🤗🤩
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 .
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 . 🤣🤣🤣