Question

the roots of QE ,x+1/x=3,x≠0​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

$x + \frac{1}{x} = 3$

${x}^{2} + 1 = 3x$

${x}^{2} - 3x + 1 = 0$

Using Shridhracharya formula,

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

$x = \frac{ - ( -3) \pm \sqrt{ {( - 3)}^{2} - 4 \times 1 \times 1 } }{2}$

$x = \frac{3 \pm \sqrt{9 - 4} }{2}$

$x = \frac{3 \pm \sqrt{5} }{2}$

$x = \frac{3 + \sqrt{5} }{2} \: \: \: or \: \: \: \frac{3 - \sqrt{5} }{2}$