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If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( p(x)=4 x^{2}+3 x+7 \), then find the value of: (a) \( \frac{1}{\alpha}+\frac{1}{\beta} \)(b) \( \alpha^{2}+\beta^{2}-3 \alpha \beta \) (c) \( \frac{4}{\alpha}+\frac{4}{\beta} \) (d) \( \left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+6 \alpha \beta \) |
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Answer» α + ß = -3/4 αß = 7/4 (a) \(\frac{α+ß}{αß}=\frac{-3}7\) ⇒ \(\frac1\alpha+\frac1{\beta}\) = \(\frac{-3}7\) (b) α2 + ß2 - 3αß = (α + ß)2 - 5αß = \((\frac{-3}4)^2 - 5\times\frac74\) = \(\frac9{16}-\frac{35}4\) = \(\frac{9-140}4\) = \(\frac{-131}4\) (c) \(\frac4{\alpha}+\frac4{\beta}=\frac{-12}4\) (From (a)) (d) \((\frac{\alpha}{\beta}+\frac{\beta}{\alpha})+6\alpha\beta\) \(=\frac{\alpha^2+\beta^2 +2\alpha\beta-2\alpha \beta}{\alpha\beta}+6\alpha\beta\) \(=\frac{(\alpha^2+\beta^2)^2-2\alpha\beta}{\alpha\beta}+6\alpha\beta\) \(=\cfrac{(-\frac34)^2-2\times\frac74}{\frac74}+6\times\frac74\) \(=\cfrac{\frac{9-56}{16}}{\frac74}+\frac{21}2\) \(=\frac{-47}{7\times4}+\frac{21}2\) \(=\frac{-47+294}{28}\)\(=\frac{247}{28}\) |
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