can you help me to solve quartic equation x^4+6*x^3-5*x^2 -10*x-3=0

Question

By putting x= y-a/4=y -6/4= y-3/2 ;then substituting and expanding  and adding I had

y^4-6y^3+54/4*y^2-108/8*y+6y^3-54/2*y^2+162/4*y -162/8 -5y^2 +15y -45/4 -10y +15

-3=0 , then by adding and combing like terms  I have y^4 +(- 37/2*y^2) +32y^2 +(-231/16)=0 . then y^4 -37/2*y^2 +32y^2 -231/16=0 ;therefore  e= -37/2 ,f =32 ,

g =-231/16 . Having  previously derived,h^6 +2*e*h^4 +(e^2-4*g)*h^2 -f ^2=0 .

how do I get from this cubic to h^6 +(-13)*h^4 +46*h^2 -1=0 and deriving that  h^2

equals 0.0218741248 ?

Answer 1

1 1/2

1.5

3/2

=x

I think

Answer 2

Four roots as follows: x1=1/2+sqrt(5)/2, x2=1/2-sqrt(5)/2, x3=-7/2+sqrt(37)/2, x4=-7/2-sqrt(37)/2

 

Basically, you can factor this quartic into the product of two quadratics.

 

(x^2-x-1)*(x^2+7x+3)

Answer 3

Look for 0's they cancel out ... looking at equation 3-3=0 so 3s are gone from equation. Multiplication 1st 5x2=10 now add and subtract 4+6=10 .... only numbers remain are 10 10 10 ... 10 greater than -10= 0 0x10=0 in other words correspondence of 10s cancel each other out