Question

If alpha and beta are zeroes of x2-5x + 6 then find the value of 1/alpha2 + 1/beta2

Answer 1

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Answer 2

Answer:

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\frac{13}{36}$

Step-by-step explanation:

$Given\: \alpha \:and\: \beta \\are\: zeroes\: of\: x^{2}-5x+6$

Compare x²-5x+6 with ax²+bx+c , we get

a = 1 , b = -5 , c = 6

$i) Sum\: of \:the\: zeroes\\=\frac{-b}{a}$

$\implies \alpha+\beta = \frac{-(-5)}{1} \\=5 ---(1)$

$ii) Product \:of\: the\: zeroes\\=\frac{c}{a}$

$\implies \alpha\beta = \frac{6}{1} \\=6---(2)$

$iii) \alpha^{2}+\beta^{2}\\=(\alpha+\beta)^{2}-2\alpha\beta \\=5^{2}-2\times 6\\=25-12\\=13---(3)$

$Now, \\\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}\\=\frac{(\beta^{2}+\alpha^{2})}{(\alpha\beta)^{2}}$

=$\frac{13}{6^{2}}\\=\frac{13}{36}$

Therefore,

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\frac{13}{36}$

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