Can you Pass Engineering Admission Exam from Harvard University ?

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Күн бұрын

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@raghvendrasingh1289 4 күн бұрын

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.

@superacademy247 4 күн бұрын

That's a great solution for this specific problem! 🤩💪This is a clever approach! Thanks for sharing your perspective. 🙏💯

@билал-ж2к 4 күн бұрын

Великолепное решение!!!❤

@kareolaussen819 4 күн бұрын

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.

@superacademy247 3 күн бұрын

Thanks for your shortcut 🙏👏💡😎💕💯

@mrflibble5717 3 күн бұрын

Excellent!

@superacademy247 3 күн бұрын

I'm glad you enjoyed the video! 😁💯💕😎🙏Thanks for the kind words! 😊

@ShriH-d1o 4 күн бұрын

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)

@davidseed2939 3 күн бұрын

can every quartic with no x³ term be expressed as the product of matching quadratics

@ShriH-d1o 3 күн бұрын

@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 .

@Jjsksskyndo 2 күн бұрын

182.19

@matheusivanmedinaramirez772 2 күн бұрын

A can't understand the step on 4:28, explain me please!

@superacademy247 2 күн бұрын

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💪✅💕💯🤗🤩