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

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

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


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

we start with

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

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.

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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 $u^3$ & $v^3$ are the solution of the quadratic equation

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

This equation has the solutions

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

$u$ & $v$ 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 $usogorsk.combbC$:$$left 1 , sqrt3,iover2-1over2 , -sqrt3,iover2-1over2 ight ag16$$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

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