Find the zero of the polynomial t² -15 and verify the relationship be

1.	Find the zero of the polynomial t² -15 and verify the relationship between the zero and coefficient
Answer» $\rule{250}3$ $\huge\sf\red{Given\::}\\$ $\sf{Polynomial\: : \: t^2 \:-15}$ $\leadsto\:\:\sf t^2 - (\sqrt{15})^2$ $\rule{250}3$ $\huge\sf\pink{Solution\::}$ $\small{\underline{\boxed{\sf{ a^2 - b^2 \:=\: (a \:-\:b) (a\:+\:b)}}}}$ $\leadsto\:\sf (t - \sqrt{15}) (t + \sqrt{15}) = 0$ $\leadsto\:\sf t = \sqrt{15} \:\:\:\; t = - \sqrt{15}$ $\leadsto\:\sf \alpha = \sqrt{15} \:\:\: \: \beta = - \sqrt{15}$ $\sf Zeroes\; of \; the \: Given \; polynomial \; =\; \sqrt{15} \:\:\:\; -\sqrt{15}$ $\rule{250}3$ $\sf\green{Sum\: of \: Zeroes\::}$ $\leadsto\:\sf \alpha + \beta = -\dfrac{b}{a}$ $\leadsto\:\sf \sqrt{15} - \sqrt{15} = -\dfrac{0}{1}$ $\leadsto\:\sf 0 = 0$ $\rule{250}3$ $\sf\purple{Product\: of \; Zeroes\::}$ $\leadsto\:\sf \alpha \;\beta = \dfrac{c}{a}$ $\leadsto\:\sf (\sqrt{15}) (-\sqrt{15}) = \dfrac{-15}{1}$ $\leadsto\:\sf -(\sqrt{15})^2 = \dfrac{-15}{1}$ $\leadsto\:\sf -15 = -15$ $\large\bold{\underline{\sf{\red{Hence\: verified\:!!}}}}$ $\rule{250}3$