Question

solve 2(x² + 1/x²)-5(x + 1/x)+6 = 0​

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: solve \: for : \: value(s) \: of \: x \\ \\ given \: biquadratic \: equation \: is \\ 2(x {}^{2} + \frac{1}{x {}^{2} } \: ) - 5(x + \frac{1}{x} \: ) + 6 = 0 \\ \\ 2[(x + \frac{1}{x} ) {}^{2} - 2x. \frac{1}{x} ] - 5(x + \frac{1}{x} ) + 6 =0 \\ \\ 2[(x + \frac{1}{x} ) {}^{2} - 2] - 5(x + \frac{1}{x} ) + 6 = 0 \\ \\ let \: here \\ x + \frac{1}{x} = t \\ \\ 2(t {}^{2} - 2) - 5t + 6 = 0 \\ 2t {}^{2} - 4 - 5t + 6 = 0 \\ 2t {}^{2} - 5t + 2 = 0 \\ 2t {}^{2} - 4t - t + 2 = 0 \\ 2t(t - 2) - 1(t - 2) = 0 \\ \\ t - 2 = 0 \: \: or \: \: 2t - 1 = 0 \\ t = 2 \: \: or \: \: t = \frac{1}{2}$

$thus \: then \: resubstituting \\ we \: get \\ \\ x + \frac{1}{x} = 2 \: \: or \: \: x + \frac{1}{x} = \frac{1}{2} \\ \\ considering \: first \\ x + \frac{1}{x} = 2 \\ \\ x {}^{2} + 1 = 2x \\ x {}^{2} - 2x + 1 = 0 \\ (x - 1) {}^{2} = 0 \\ (x - 1)(x - 1) = 0 \\ \\ x = 1$

$now \: considering \: \\ x + \frac{1}{x} = \frac{1}{2} \\ \\ \frac{x {}^{2} + 1 }{x} = \frac{1}{2} \\ \\ 2(x {}^{2} + 1) = x \\ 2x {}^{2} + 2 = x \\ 2x {}^{2} - x + 2 = 0 \\ \\ comparing \: it \: with \: ax {}^{2} + bx + c = 0 \\ we \: get \\ a = 2 \\ b = - 1 \\ c = 2 \\ \\ we \: know \: that \\ Δ = b {}^{2} - 4ac \\ = ( - 1) {}^{2} - 4 \times 2 \times 2 \\ = 1 - 16 \\ Δ= - 15 \\ \\ now \: by \: using \\ \: shreedharacharya \: method \\ we \: get \\ \\ x = \frac{ - b± \sqrt{b {}^{2} - 4ac} } {2a} \\ \\ = \frac{ - ( - 1)± \sqrt{ - 15} }{2 \times 2} \\ \\ = \frac{ - 1 ± \sqrt{15} \times \sqrt{ - 1} }{4} \\ \\ x = \frac{1± \sqrt{15} \: i }{4}$