What are the roots of x^3-4*x^2+9x-36 = 0

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find the three solutions to the given equation, hence, you need to group the terms such that:

`(x^3 - 4x^2) + (9x - 36) = 0`

Notice that you may factor out `x^2`  in the group `x^3 - 4x^2`  and you may factor out 9 in the group `9x - 36` , such that:

`x^2(x - 4) + 9(x - 4) = 0`

You may factor out `(x - 4)`  such that:

`(x - 4)(x^2 + 9) = 0`

You need to solve for x the following equations such that:

`{(x - 4 = 0),(x^2 + 9 = 0):}` `=> {(x = 4),(x^2 = -9):}`

You need to use the complex number theory to solve the equation `x^2 = -9`  such that:

`x^2 = -9=> x_(1,2) = +-sqrt(-9)`

Since `sqrt(-1) = i`  yields:

`x_(1,2) = +-3i`

Hence, evaluating the roots to the given cubic equation yields one real root `x = 4`  and two complex conjugate roots `x_(1,2) = +-3i.`

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