If alpha and beeta are the …
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Sia ? 6 years, 6 months ago
Since, α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x - 2.
a=6, b=1, c=-2
sum of zeros =α+β={tex}-\frac ba{/tex}{tex}= \frac{{ - 1}}{6}{/tex}
Product of the zeroes = αβ={tex}\frac ca{/tex} {tex} = \frac{{ - 1}}{3}{/tex}
Now,
{tex}\frac{\mathrm\alpha}{\mathrm\beta}+\frac{\mathrm\beta}{\mathrm\alpha}=\frac{\mathrm\alpha^2+\mathrm\beta^2}{\mathrm{αβ}}=\frac{(\mathrm\alpha+\mathrm\beta)^2-2\mathrm{αβ}}{\mathrm{αβ}}{/tex}
{tex}=\frac{\left(-{\displaystyle\frac16}\right)^2-2(-{\displaystyle\frac13})}{\displaystyle-\frac13}=\frac{\displaystyle\frac1{36}+\frac23}{\displaystyle-\frac13}=-\frac{\displaystyle\frac{75}{36}}{\displaystyle\frac13}=-\frac{75}{36}\times\frac31=-\frac{25}4{/tex}
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