Question

Find the zeroes of the quadratic polynomial 6x square-3x-7 x and verify the relationship between the zeroes and coefficient of the polynomial

Answer 1

→ HEY THERE!!→

$\bold{\huge{SOLUTION:-}}$



$\bf{Given:-}$

→ Quadratic Equation: 6x²-7x-3


$\bf{Method \: \: of \: \: Solution:-}$



→ Quadratic Equation: 6x²-7x-3

→ 6x²-7x-3

→ 6x²+2x-9x -3

→ 2x(3x+1)-3(3x+1)

→(3x+1)(2x-1)


For quadratic equation f(x) = (3x+1)(2x-1)


f(x) = (3x+1)(2x-1) → 0

→ 3x+1 = 0

→ x = -1/3

→ (2x-1) = 0

→ x = 1/2


Now, Verify the Relationship between the Zeroes and Coefficient of the Polynomial!



Sum of Zeroes = -b/a → -(Coefficient of x)/coefficient of x²

→ a+b = -b/a

-1/3 + 1/2 = 1/6 → -(Coefficient of x)/coefficient of x²



Now,



Product of Zeroes = c/a → (Constant term )/coefficient of x²


→ -1/3 × 1/2 = -1/6 → (Constant term )/coefficient of x²




$\bf{Hence,\: \: It's \: \: Proved}$

Answer 2

Quadratic Equation: 6x²-7x-3

6x²-7x-3

6x²+2x-9x -3

2x(3x+1)-3(3x+1)

(3x+1)(2x-1)

Verify the Relationship between the Zeroes and Coefficient of the Polynomial!



Sum of Zeroes =) -b/a -(Coefficient of x)/coefficient of x²

a+b = -b/a

-1/3 + 1/2 =) 1/6 -(Coefficient of x)/coefficient of x²



Now,



Product of Zeroes = c/a (Constant term )/coefficient of x²


c/a = -1/6 (Constant term )/(coefficient of x²)