Question

Find the zero of the polynomial t² -15 and verify the relationship between the zero and coefficient​

Answer 1

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$\huge\sf\red{Given\::}$

$\sf{Polynomial\: : \: t^2 \:-15}$

$\leadsto\:\:\sf t^2 - (\sqrt{15})^2$

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$\huge\sf\pink{Solution\::}$

$\small{\underline{\boxed{\sf{ a^2 - b^2 \:=\: (a \:-\:b) (a\:+\:b)}}}}$

$\leadsto\:\sf (t - \sqrt{15}) (t + \sqrt{15}) = 0$

$\leadsto\:\sf t = \sqrt{15} \:\:\:\; t = - \sqrt{15}$

$\leadsto\:\sf \alpha = \sqrt{15} \:\:\: \: \beta = - \sqrt{15}$

$\sf Zeroes\; of \; the \: Given \; polynomial \; =\; \sqrt{15} \:\:\:\; -\sqrt{15}$

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$\sf\green{Sum\: of \: Zeroes\::}$

$\leadsto\:\sf \alpha + \beta = -\dfrac{b}{a}$

$\leadsto\:\sf \sqrt{15} - \sqrt{15} = -\dfrac{0}{1}$

$\leadsto\:\sf 0 = 0$

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$\sf\purple{Product\: of \; Zeroes\::}$

$\leadsto\:\sf \alpha \;\beta = \dfrac{c}{a}$

$\leadsto\:\sf (\sqrt{15}) (-\sqrt{15}) = \dfrac{-15}{1}$

$\leadsto\:\sf -(\sqrt{15})^2 = \dfrac{-15}{1}$

$\leadsto\:\sf -15 = -15$

$\large\bold{\underline{\sf{\red{Hence\: verified\:!!}}}}$

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