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Tcecarenko [31]
3 years ago
12

How do I solve this?​

Mathematics
1 answer:
ch4aika [34]3 years ago
4 0

Answer:

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

Step-by-step explanation:

1 +  \frac{1}{ {x}^{2} }  =  \frac{3}{x}  \\  \\  \frac{ {x}^{2} + 1 }{ {x}^{2} }  =  \frac{3}{x}  \\  \\  {x}^{2}  + 1 =  \frac{3 {x}^{2} }{x}  \\  \\  {x}^{2}  + 1 = 3x \\  \\  {x}^{2}  - 3x + 1 = 0 \\  equating \: it \: with \\ a {x}^{2}  + bx + c = 0 \\ a  = 1 \\ b =  - 3 \\ c = 1 \\  \\ x =  \frac{ - b \pm \sqrt{ {b}^{2}  - 4ac} }{2a}  \\  \\  x =  \frac{ - ( - 3) \pm \sqrt{ {( - 3)}^{2}  - 4 \times 1 \times 1} }{2 \times 1}  \\  \\ x =  \frac{ 3 \pm \sqrt{ {9  - 4}} }{2}  \\  \\ x =  \frac{ 3 \pm \sqrt{ {5}} }{2}  \\  \\ x =  \frac{3}{2}  +  \frac{ \sqrt{5} }{2} ; \:  \: x = \frac{3}{2}   -  \frac{ \sqrt{5} }{2}

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