analytically? That is, without using numerical methods lớn attain an approximate solution.

Bạn đang xem: The art and science of change ringing

By the rational root test, this equation doesn"t have any rational roots.

So you have to use the general method for solving cubic equations.

Using Cardano"s method (and a cas to vì chưng the calculations):

\$\$x^3+x^2-4=0 ag1\$\$

and substitute \$\$x=t-1over3 ag2\$\$to get\$\$t^3-tover3-106over27=0 ag3\$\$

It is always possible khổng lồ find a substitution \$x=t+a\$ such that the coefficient of \$x^2\$ vanishes.

Xem thêm: Cách Đổi Hình Nền Slide Trong Powerpoint Để Hấp Dẫn Hơn

Now we mix \$\$t=v+u ag4\$\$and get

\$\$v^3+left(v+u ight),left(9,u,v-1 ight)over3+u^3-106 over27=0 ag5\$\$

We impose the additional condition that \$\$u,v=1over9 ag6\$\$

and get

\$\$v^3+u^3=106over27 ag7\$\$

and

\$\$u^3,v^3=1over729 ag8\$\$

So \$u^3\$ và \$v^3\$ are the solution of the quadratic equation

\$\$z^2-106,zover27+1over729=0 ag9\$\$

This equation has the solutions

\$\$left -6,sqrt78-53over27 , 6,sqrt78+53 over27 ight ag10\$\$

\$u\$ và \$v\$ are interchangeable, we set

\$\$u^3=-6,sqrt78-53over27 ag11\$\$\$\$v^3=6,sqrt78+53 over27 ag12\$\$

Both values have a real cubic root. We can set

\$\$u=sqrt<3>-6,sqrt78-53over27 ag13\$\$\$\$v=sqrt<3>6,sqrt78+53 over27 ag14\$\$

because these solution pair satisfies \$(1)\$.

From this we get \$\$x=sqrt<3>-6,sqrt78-53over27 + sqrt<3>6,sqrt78+53 over27 +1/3 ag15\$\$

\$1\$ has three roots in \$firmitebg.combbC\$:\$\$left 1 , sqrt3,iover2-1over2 , -sqrt3,iover2-1over2 ight ag16\$\$So the two other soltuions we construct from

\$\$u=(sqrt<3>-6,sqrt78-53over27) (sqrt3,iover2-1over2) ag17\$\$\$\$v=(sqrt<3>6,sqrt78+53 over27) (-sqrt3,iover2-1over2) ag18\$\$

and

\$\$u=(sqrt<3>-6,sqrt78-53over27) (-sqrt3,iover2-1over2) ag19\$\$\$\$v=(sqrt<3>6,sqrt78+53 over27) (sqrt3,iover2-1over2) ag20\$\$