If \(\alpha\) and \(\beta\) are the roots of \(x^{2} + x - 2 = 0\), find the value of \(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}}\).
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Correct Answer: Option A
Explanation:
Given, \(x^{2} + x - 2 = 0\), a = 1, b = 1 and c = -2.
\(\alpha + \beta = \frac{-b}{a} = \frac{-1}{1} = -1\)
\(\alpha\beta = \frac{c}{a} = \frac{-2}{1} = -2\)
\(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{\beta^{2} + \alpha^{2}}{(\alpha\beta)^{2}}\)
\(\beta^{2} + \alpha^{2} = (\alpha + \beta)^{2} - 2\alpha\beta = (-1)^{2} - 2(-2) = 1 + 4 = 5\)
\(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{5}{(-2)^{2}} = \frac{5}{4}\).
Given, \(x^{2} + x - 2 = 0\), a = 1, b = 1 and c = -2.
\(\alpha + \beta = \frac{-b}{a} = \frac{-1}{1} = -1\)
\(\alpha\beta = \frac{c}{a} = \frac{-2}{1} = -2\)
\(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{\beta^{2} + \alpha^{2}}{(\alpha\beta)^{2}}\)
\(\beta^{2} + \alpha^{2} = (\alpha + \beta)^{2} - 2\alpha\beta = (-1)^{2} - 2(-2) = 1 + 4 = 5\)
\(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{5}{(-2)^{2}} = \frac{5}{4}\).