Question

find zeros of t square minus 25 and verify relationship between zeros and coefficient

Answer 1

hope it helps
regards
Mohit Khairnar

Answer 2

${t }^{2} - 25 \\ {(t)}^{2} - {(5)}^{2} \\ (t + 5)(t - 5) = 0 \\ \\ (t + 5) = 0 \\ t = - 5 \\ \\ (t - 5) = 0 \\ t = 5 \\ \\ \alpha = - 5 \\ \beta = 5 \\ \\ (a) \: \alpha + \beta = \frac{ - b}{a} \\ 5 - 5 = - \frac{0}{1} \\ 0 = 0 \\ \\ (b) \: \alpha \beta = \frac{c}{a} \\ 5 \times ( - 5) = \frac{ - 25}{1} \\ - 25 = - 25 \\ verified$
very simple...