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    Subject

    Bank Exam

    Topic

    IFIC Bank Ltd. - Trainee Assistant Officer - 2018

    If x2- 3x + 1 = 0, What is the value of x2  −1x2  {x^{2\;}} - \frac{1}{{{x^{2\;}}}} ?

    ক)
    \(4\sqrt 3 \)
    খ)
    \(3\sqrt 5 \)
    গ)
    \(4\sqrt 5 \)
    ঘ)
    \(2\sqrt 3 \)

    Explanation

    Solution: আমাদেরকে বের করতে হবে, x2  −1x2  {x^{2\;}} - \frac{1}{{{x^{2\;}}}} = (x+1x)(x−1x)(x + \frac{1}{x})(x - \frac{1}{x}) .......... (1) অর্থাৎ আমরা (x+1x)(x + \frac{1}{x}) এবং (x−1x)(x - \frac{1}{x}) এর মান নিচের সমীকরণের সাহায্যে বের করবো । x 2 - 3x + 1 = 0 => x2x−3xx+1x=0x\frac{{{x^2}}}{x} - \frac{{3x}}{x} + \frac{1}{x} = \frac{0}{x}         [x দ্বারা ভাগ করে ] => x−3+1x=0x - 3 + \frac{1}{x} = 0 => x+1x=3x + \frac{1}{x} = 3                  [পক্ষান্তর করে ] ......... (2) =>  (x+1x)2  =(3)2  {(x + \frac{1}{x})^{2\;}} = {(3)^{2\;}}         [ বর্গ করে ] => (x−1x)2  +4×x×1x=9{(x - \frac{1}{x})^{2\;}} + 4 \times x \times \frac{1}{x} = 9 => (x−1x)2  +4=9{(x - \frac{1}{x})^{2\;}} + 4 = 9 => (x−1x)2  =9−4{(x - \frac{1}{x})^{2\;}} = 9 - 4 => (x−1x)2  =5{(x - \frac{1}{x})^{2\;}} = 5 (x−1x)=5(x - \frac{1}{x}) = \sqrt 5         [বর্গমূল করে ] .......... (3) অর্থাৎ আমরা পেলাম (x+1x)=3(x + \frac{1}{x}) = 3 এবং (x−1x)=5(x - \frac{1}{x}) = \sqrt 5 x2−1x2=(x+1x)(x−1x)=3×5 =35{x^2} - \frac{1}{{{x^2}}} = (x + \frac{1}{x})(x - \frac{1}{x}) = 3 \times \sqrt 5  = 3\sqrt 5

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