The zeroes of quadratic polynomial x2-x-4 …
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Sia ? 6 years, 6 months ago
F(x)=x2-x-4
here a=1,b=-1,c=-4
Now {tex}\mathrm\alpha\;\mathrm{and}\;\mathrm\beta{/tex} are zeros of f(x)
Sum of zeroes = α + β {tex}=-\frac ba{/tex}=-{tex}\frac{-1}{1}{/tex}=1
Product of the zeroes = αβ = {tex}\frac ca{/tex}={tex}\frac{-4}{1}=-4{/tex}
{tex}\frac1{\mathrm\alpha}+\frac1{\mathrm\beta}-\mathrm{αβ}=\frac{\mathrm\alpha+\mathrm\beta-\mathrm\alpha^2\mathrm\beta^2}{\mathrm{αβ}}=\frac{1-(-4)^2}{-4}{/tex}
{tex}=\frac{1-16}{-4}=\frac{-15}{-4}=\frac{15}4{/tex}
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