# Answer on Algebra Question for Inga

Question #4515

How to get exact solutions of equation x^3+3x^2-1=0.

The solution can't be decimal approximation, it should be analytical solution.

The solution can't be decimal approximation, it should be analytical solution.

Expert's answer

x^3+3x^2-1=0

We use here trigonometric Viete formula

for equation written in form x^3+a x^2+b x+c=0.

Q=(a^2-3b)/9

R=(2a^3-9ab+27c)/54

S=Q^3-R^2

For our equation Q=(3^2-3*0)/9=1

R=27/54=1/2

S=3/4

fi=1/3*arccos(R/Q

x1=-2sqrt(Q) cos fi-a/3

x2=-2sqrt(Q) cos( fi+2 pi/3)-a/3

x3=-2sqrt(Q) cos( fi-2 pi/3)-a/3

for our data

fi=1/3 arccos(0.5)

x1=-2 cos(1/3*arccos(0.5))-1

x2=-2 cos(1/3*arccos(0.5)+2 pi/3)-1

x3=-2 cos(1/3*arccos(0.5)-2 pi/3)-1

where x1, x2, x3 - the roots of our equation

We use here trigonometric Viete formula

for equation written in form x^3+a x^2+b x+c=0.

Q=(a^2-3b)/9

R=(2a^3-9ab+27c)/54

S=Q^3-R^2

For our equation Q=(3^2-3*0)/9=1

R=27/54=1/2

S=3/4

fi=1/3*arccos(R/Q

^{3/2})x1=-2sqrt(Q) cos fi-a/3

x2=-2sqrt(Q) cos( fi+2 pi/3)-a/3

x3=-2sqrt(Q) cos( fi-2 pi/3)-a/3

for our data

fi=1/3 arccos(0.5)

x1=-2 cos(1/3*arccos(0.5))-1

x2=-2 cos(1/3*arccos(0.5)+2 pi/3)-1

x3=-2 cos(1/3*arccos(0.5)-2 pi/3)-1

where x1, x2, x3 - the roots of our equation

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