Question

find the zeros of 6s square + s - 12 and verify the relationship between zeroes and coefficients​

Answer 1

Solution:

Given Quadratic Polynomial → 6s² + s - 12

Finding zeros by factorization.

6s² + s - 12 = 0

⇒ 6s² + 9s - 8s - 12 = 0

⇒ 3s (2s + 3) - 4(2s + 3) = 0

⇒ (3s - 4) (2s + 3) = 0

⇒ s = 4/3 or -3/2

∴ The zeros of the Quadratic polynomial are:

• 4/3
• -3/2

Finding the sum of zeroes:

$\sf{Sum\:of\:zeroes=\dfrac{-b}{a}}$

$\implies \sf{\dfrac{4}{3}-\dfrac{3}{2}=\dfrac{-1}{6}}$

$\dashrightarrow \: \: \sf{\dfrac{-1}{6}=\dfrac{-1}{6}}$

Hence the sum of zeroes verified

Finding the product of zeroes:

$\sf{Product \: of \: zeroes = \dfrac{c}{a}}$

$\implies \sf{\dfrac{4}{3}\times\dfrac{-3}{2}=\dfrac{-12}{6}}$

$\implies \sf{\dfrac{4}{\not{3}}\times\dfrac{-\not{3}}{2}=\dfrac{-12}{6}}$

$\dashrightarrow \: \: \sf{\dfrac{-12}{6}=\dfrac{-12}{6}}$

Hence the product of zeros is also verified.