Question

solve the cubic equation x³-6x-9=0 by Cardan's method.​

Answer 1

Answer⤵️

Given:

$\rm{x}^{3} - 6x + 9 = 0$

$\rm{ let \: x = - 3 } \\ \rm{ - 27 + 18 + 9 = 0}$

$\rm{hence \: x - 3 \: is \: a \: factor \: of \: {x}^{3} - 6x + 9 = 0}$

Thus the equation can be written as

$\rm{ = (x + 3)( {x}^{2} - 3x + 3) = 0} \\ \rm{{x}^{2} - 3x + 3 = 0 \: and \: x + 3 = 0} \\ \rm{consider \: {x}^{2} - 3x + 3 = 0} \\ \rm{ {b}^{2} - 4ac} \\ \rm{ = 9 - 12} \\ \rm{ = 3}$

$\rm{hence \: {b}^{2} - 4ac < 0. \: hence \: imaginary \: roots.}$

$\rm{Therefore, \: the \: only \: real \: root \: is \: x = - 3.}$

hope this helps!

Answer 2

Answer:

Step-by-step explanation: The other answer is not Cardano’s method

I am going to solve it Cardano’s method.

x^3-6x-9=0

Let x=u+v

(u+v)^3-6(u+v)-9=0

u^3+v^3+3uv(u+v)-6(u+v)-9=0

u^3+v^3+(u+v)(3uv-6)-9=0

Now we want to make the middle mess vanish,so we set

3uv-6=0

uv=2

(uv)^3=8

This will also give us u^3+v^3=9

Now lets make a quadratic equation in t with roots u^3 and v^3

t^2-(sum of roots)t+(product of roots)=0

t^2-9t+8=0

t^2-t-8t+8=0

t(t-1)-8(t-1)=0

(t-1)(t-8)=0

t=1,8
u and v are interchangeable.

So we are going to let u^3=1 and v^3=8

This means u=1 and v=2

Therefore, x=u+v=1+2=3

Divide the cubic by x-3 and solve the remaining quadratic to get the other two roots.