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Earlier discussion
1 older comment kept for context.
Oluwaseun Adeyemi
Generally, quadratic equations are given as \(ax^2 + bx + c = 0\)
Comparing the given equation to the above, gives \(a = 1, b = -10\) and \(c = 2\)
If \(\alpha\) and \(\beta\) are the roots of an equation, then \(\alpha + \beta = \cfrac{-b}{a}\) and \(\alpha \beta = \cfrac{c}{a}\)
\(\cfrac{1}{\alpha^2}+\cfrac{1}{\beta^2} = \cfrac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \cfrac{(\alpha + \beta)^2 - 2\alpha \beta}{(\alpha \beta)^2}\)
\(\alpha + \beta = \cfrac{-(-10)}{1}\) and \(\alpha \beta = \cfrac{2}{1}\)
\(\alpha + \beta = 10\) and \(\alpha \beta = 2\)
\(\cfrac{1}{\alpha^2}+\cfrac{1}{\beta^2} = \cfrac{(\alpha + \beta)^2 - 2\alpha \beta}{(\alpha \beta)^2} = \cfrac{10^2 - 2(2)}{2^2}\)
\(\cfrac{100-4}{4}=\cfrac{96}{4}=24\)