X 3 4x 2 x 1 x 2 3

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X3-4x2-x+1=(x-2)3 ansk fast

Answered by admin @


Answer :

{x}^{3} - 4 {x}^{2} - x + 1 = {(x - 2)}^{3} \\ \\ = > {x}^{3 } - 4 {x}^{2} - x + 1 = {x}^{3} - 3 {(x)}^{2}(2) + 3(x) {(2)}^{2} - {(2)}^{3} \\ \\ = > {x}^{3} - 4 {x}^{2} - x + 1 = {x}^{3} - 6 {x}^{2} + 12x - 8 \\ \\ = > 2 {x}^{2} - 13x + 9 = 0

Now, By Shridharacharya Formula :

Here a = 2, b = -13 and c = 9

x = \frac{ - {b} \binom{ + }{ - } \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ = > x = \frac{ - ( - 13) \binom{ + }{ - } \sqrt{ { ( - 13)}^{2} - 4(2)(9) } }{2 \times 2} \\ \\ = > x = \frac{13 \binom{ + }{ - } \sqrt{169 - 72} }{4} \\ \\ = > x = \frac{13 \binom{ + }{ - } \sqrt{97} }{4} \\ \\ = > x = \frac{13 + \sqrt{97} }{4} ,\: \frac{13 - \sqrt{97} }{4}

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