Harvard University Admission Interview Tricks You Should Know!✍️🖋️📘💙

[@에스피-z2g]

x^4+4x^3+2x^2-4x
+1=0
(x-1)(x+1)(x^2+4x-1)=0
x=1,-1,-2+rt5,
-2-rt5

[@1942Reinaldo]

There is a mistake in your approach. You already have found it? +2x^2 and not -2x^2 !!!

[@에스피-z2g]

@1942Reinaldo
Oh, my mistake.
Thanks for your
advice. Will try to find the right answer.

[@prollysine]

/:x^2 , x^2+4/x+2-4/x+1/x^2=0 , (x+1/x)^2=x^2+1/x^2+2 , (x-1/x)^2=x^2+1/x^2-2 , let u=x-1/x , we get , U^2{+2}+4u+2=0 ,
u^2+4u+4=0 , u= -2 , x-1/x= -2 , x^2+2x-1=0 , x=(-2+/-V8)/2 , x= -1+/-V2 , x= -1+V2 , -1-V2 ,

[@ChavoMysterio]

(x⁴+4x³+2x²)/(4x-1)=1
Domain: x≠¼
x⁴+4x³+2x²=4x-1
x⁴+4x³+2x²-4x+1=0
x²+4x+2-(4/x)+[1/(x²)]=0
x²+[1/(x²)]+4x-(4/x)+2=0
[x-(1/x)]²+2+4[x-(1/x)]+2=0
[x-(1/x)]²+4[1-(1/x)]+4=0
Let y=[x-(1/x)]
y²+4y+4=0
(y+2)²=0
y+2=0
y=-2
[x-(1/x)]=-2
x²-1=-2x
x²+2x-1=0
x²+2x+1=2
(x+1)²=2
|x+1|=√2
x+1=±√2
x=-1±√2 ❤❤