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

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beckden | High School Teacher | (Level 1) Educator

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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.

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beckden | High School Teacher | (Level 1) Educator

Posted on

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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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).`

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erheol | eNotes Newbie

Posted on

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.

 

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erheol | eNotes Newbie

Posted on

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?

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