Question

(12x-1)(6x-1)(4x-1)(3x-1) =5
Solve for x

Answer 1

Answer:

$x = \frac{1}{2} \: or \: x = \frac{ - 1}{12}$

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Step-by-step explanation:

$(12x - 1)(6x - 1)(4x - 1)(3x - 1) =5 \\ (2 \times 3 \times 4)(12x - 1)(6x - 1)(4x - 1)(3x - 1) =5(2 \times 3 \times 4) \\ (12x - 1)2(6x - 1)3(4x - 1)4(3x - 1) =5(2 \times 3 \times 4 \\ (12x - 1)(12x - 2)(12x - 3)(12x - 4) =(5)(24) \\ (12x - 1)(12x - 1 - 1)(12x - 1 - 2)(12x - 1 - 3) =(5)(24) \\ Now \: lets \: take \: (12x-1) as s \\ therefore \\ (s)(s - 1)(s - 2)(s - 3) = 120 \\ (s)(s - 3)(s - 1)(s - 2) = 120 \\ ( {s}^{2} - 3s)( {s}^{2} - 3s + 2) = 120 \\ Now \: lets \: take \: ( {s}^{2} - 3s ) as r \\ therefore \\ (r)(r + 2) = 120 \\ {r}^{2} + 2r = 120 \\ Adding 1 on both sides \\ {r}^{2} + 2r + 1 = 120 + 1 \\ {(r + 1)}^{2} = 121 \\ {(r + 1)}^{2} = {11}^{2} \\ (r + 1) = 11 \: or \: (r + 1) = ( - 11) \\ r = 10 \: or \: r = ( - 12) \\ As \: r = ({s}^{2} - 3s) \\ ({s}^{2} - 3s) = 10 \: or \: ({s}^{2} - 3s) = ( - 12) \\ Now \: lets \: solve \: for \: ({s}^{2} - 3s) = ( - 12) : \\ Therefore \: ({s}^{2} - 3s) - ( - 12) = 0 \\ {s}^{2} - 3s + 12 = 0 \\ Solving by Quadratic formula i.e. \\ x = \frac{( - b) \frac{ + }{ - } \sqrt{ {b}^{2} - 4ac } }{2a} \\ we \: get \: \\ s = \frac{( 3) \frac{ + }{ - } \sqrt{ - 27} }{2} \\ Here \: we \: get \: negative \: root \: in \: solution \\ Neglecting \: negative \: root \: as \: solution \: we \: solve \: for \: another \: value \: which \: is: \\ ({s}^{2} - 3s) = 10 \\ Hence: \\ {s}^{2} - 3s - 10 = 0 \\ (s - 5)(s + 2) = 0 \\ s = (5) \: or \: s = ( - 2) \\ As \: s = (12x - 1) \: we \: get \\ 12x - 1 = 5 \: or \: 12x - 1 = ( - 2) \\ x = \frac{6}{12} \: or \: x = \frac{ - 1}{12} \\ x = \frac{1}{2} \: or \: x = \frac{ - 1}{12} </p><p> \\ \\ \\ \\ If \: we \: consider \: the \: negative \: root \: as \: solution \: of \: s \: by \: taking \: imaginary \: number \: 'i' \: where \: i = \sqrt{( - 1)} \: and \: go forward.... \\ s = \frac{( 3) \frac{ + }{ - } \sqrt{ - 27} }{2} \\ s = \frac{( 3) \frac{ + }{ - } \sqrt{ 27}i }{2} \\ s = \frac{3 + \sqrt{27}i }{2} \: or \: s = \frac{3 - \sqrt{27}i }{2} \\ As \: s = (12 - 1) \: \: \: we \: get \\ 12x - 1 = \frac{3 + \sqrt{27}i }{2} \: or \: 12x - 1 = \frac{3 - \sqrt{27}i }{2} \\ x = \frac{5 + 3 \sqrt{3}i }{24} \: or \: x = \frac{5 - 3 \sqrt{3}i }{24} \\ I \: tried \: substituting \: these \: values \: of \: imaginary \: roots \: in \: equation. \: It \: does \: not \: satisfy \: the \: equation. \\ Hence \: we \: can \: cancel \: out \: these \: imaginary \: roots! \\ \\ Hence \: solution \: of \: given \: equation \: is \\ x = \frac{1}{2} \: or \: x = \frac{ - 1}{12}$