Solving a Tricky Polynomial Equation || Tricky Solution of a Cubic Equation

Описание: Hello everyone. In this video, we have to solve this equation.

Let we move toward the solution. Given is this equation and we can write this term as x^3, so x^3 + x = 10. Taking 10 towards left, we will get x^3+x-10=0. We can write this 10 as 8 + 2 and if we open this bracket we will get x^3, so x^3 + x-8-2 = 0. So we can write this 8 as 2^3 and rearranging. We know that a^3-b^3 = (a - b)( a²+ ab + b²) and from here x^3-2^3 that will be equal to (x - 2)(x² + 2 x + 2²) and plus this term x -2 is here. Taking (x - 2) common, we will get x -2 and here x² + 2x + 4 and from here we will get 1. So the final expression is this one. If a product of two term is equal to zero then one of them must be equal to 0. So either x - 2 is equal to 0 or x² + 2x + 5 is equal 0. And from this equation we will get x = 2 this is one answer of this equation. We will find other two roots of this equation from this quadratic equation. Let we take this equation and use the famous quadratic formula on this. Here a=1, b=2 and c=5. So substituting these values over here we will get x is equal to - 2 + - sqrt(-16) / 2 and the sqrt(-16) is equal to 4i. Taking 2 Common from the numerator we will get this expression. Two in the numerator and denominator will cancel out and hence other two roots are -1 + - 2i. So the list of three Roots is x = 2, -1 + 2i and -1 - 2i. That is the answer.

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