Question

If alpha and beta are the roots of equation x^2-3x+1 ,then find 1/alpha^2+1/beta^2

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

${x}^{2} - 3x + 1 = 0 \\ compare \: equation \: with \: a {x}^{2} + bx + c = 0 \\ a = 1 \: \: \: b = - 3 \: \: \: \: c = 1 \\ {b}^{2} - 4ac = {( - 3)}^{2} - 4 \times 1 \times 1 \\ = 9 - 4 = 5 \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac} }{2a} \\ x = \frac{ - ( - 3) + - \sqrt{5} }{2} \\ x = \frac{3 + - \sqrt{5} }{2} \\ x = \frac{3 + \sqrt{5} }{2} \: \: \: \: \: or \: \: \: \: \: x = \frac{3 - \sqrt{5} }{2} \\ \alpha = \frac{3 + \sqrt{5} }{2} \: \: \: \: or \: \: \: \: \beta = \frac{3 - \sqrt{5} }{2} \\ \frac{1}{ { \alpha }^{2} } + \frac{1}{ { \beta }^{2} } = \frac{1}{( {\frac{3 + \sqrt{5} }{2})}^{2} } + \frac{1}{ {(\frac{3 - \sqrt{5} }{2})}^{2} } \\ = \frac{1}{ \frac{9 + 6 \sqrt{5} + 5 }{4} } + \frac{1}{ \frac{9 - 6 \sqrt{5} + 5}{4} } \\ = \frac{4}{9 + 6 \sqrt{5} + 5} + \frac{4}{9 - 6 \sqrt{5} + 5}$

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