Question

find the zeros of the quadratic polynomial p of x is equal to 100 x square - 81​

Answer 1

here the zeros of quadratic polynomial p of x=

$\frac{9}{10}$

Answer 2

$p(x) = 100{x}^{2} - 81$

To find zeroes of a polynomial

p(x) = 0

$p(x) = 100{x}^{2} - 81 = 0 \\ \implies {(10x)}^{2} - {(9)}^{2} = 0 \\ \implies (10x + 9) (10x - 9) = 0 \\ \implies(10x + 9) = 0 \; ;(10x - 9) =0$

$10x + 9 = 0 \\ x = \dfrac{-9}{10} \\ 10x - 9 = 0 \\x = \dfrac{9}{10}$

Hence, zeroes of $p(x) = 100{x}^{2} - 81$ are $\dfrac{-9}{10}$ and $\dfrac{9}{10}$