# solve x^3+2x^2+x-1=0 ... if possible without Cardan/Tartaglia formula

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If you have a graphing calculator like a TI-84 you can graph the equation and use the (shift trace) to find the roots. Using the Cardan formula is complicated and the answers are complicated and not easy to simplify.

There is always a real root for cubic equations. The real root is

x=0.465571231 876768026 6567312

The other roots are

x=-1. 232785615 938384013 328366-0.792551992 515447848 3258983i,

x=-1. 232785615 938384013 328366+0.792551992 515447848 3258983i

You need to group the terms `x^3 + 2x^2 + x` and to factor out x such that:

`x(x^2 + 2x + 1) - 1 = 0`

You may write the addition of terms in the brackets such that:

`x^2 + 2x + 1 = (x+1)^2`

`x(x+1)^2 - 1 = 0`

You should use the notation: x+1=y=> x = y-1

`(y - 1)*y^2 = 1 =gt y^2= 1/(y-1) =gt y = +-1/sqrt(y-1)`

`y-1gt0 =gt ygt1 =gt x+1gt1 =gt xgt0`

You need to use the derivative of equation to find the real solutions of the equation.

`(x^3 + 2x^2 + x - 1)' = 3x^2 + 4x + 1`

`3x^2 + 4x + 1 = 3x^2 + 4x + 4 - 3 = (3x^2 - 3) + (4x + 4)`

`(3x^2 - 3) + (4x + 4) = 3(x^2 - 1) + 4(x+1)`

You need to factor out (x+1) such that:

(x+1)(3x - 3 + 4) = 0 => x = -1

3x + 1 = 0 => x = `-1/3`

Since x must be positive, there is no reason to check if there are real roots of the equation between the negative values x = `-1` and x = `-1/3` .

**Hence, the solutions to the equation `x^3 + 2x^2 + x - 1` need to accomplish the condition x>0. The conclusion after evaluation of existence of the real roots is that there are no real roots of equation over the interval `(0,+oo).` **

Thanks beckden, but I should have mentionned I need to find the exact value, not an approched one (i.e. computed decimal, not sure of the exact term in english).

And still, without the Cardan formula.

Thanks for your answers.

I know there is a real root, however, I still don't know how to find it manually and without Cardan/Tartaglia formula (intended for students who aren't supposed to know it).

Any other solution?