Question

find the zeros of the quadratic polynomial 9t square - 60 + 1 and verify the relationship between the zeros and the coefficient​

Answer 1

$\bold {\huge {Answer}}$

$\bold {\underline {Question}}$

Find the zeros of the quadratic polynomial 9t^2 - 6t + 1 and verify the relationship between the zeros and the coefficient

$\bold {\underline{Answer}}$

By splitting the middle term

$9t {}^{2} - 6t + 1 = 0 \\ \\ 9 {t}^{2} -3t - 3t + 1 = 0 \\ \\ 3t(3t - 1) - 1(3t - 1) = 0 \\ \\ (3t - 1)(3t - 1) = 0 \\ \\( 3 {t}^{} - 1) {}^{2} = 0$

3t-1 = 0
t = 1/3


So these are the zeros of the given polynomial.

$\bold{\underline {What\ is\ the\ relationship\ between\ the\ zeros\ and\ the\ coefficient\ ?}}$

Here, coefficient of t^2 = 9
Coefficient of t = 6
Constant term = 1

$\bold{Sum\ of\ its\ zeros}$ = $\dfrac{2}{3}$

$\boxed{\bold{\dfrac {-b}{a} = \dfrac{2}{3} = \dfrac{Coefficient\ of\ t}{Coefficient\ of\ t^2}}}$

$\bold {Product\ of\ its\ zeros}$ = $\dfrac {1}{9}$

$\boxed{\bold{\dfrac {c}{a} = \dfrac {1}{9} = \dfrac {Constant\ term}{Coefficient\ of\ t^2}}}$

$\bold {\underline {Hence\ verified}}$