Question

Find the zeroes of the polynomial 6u²-3-7u and verify the relationship
between the zeroes and co-efficients.​

Answer 1

Step-by-step explanation:

6u² - 3 - 7u

6u² - 7u - 3

By Quadratic formula

D = b² - 4ac

= (-7)² - 4(6)(-3)

= 49 + 72

= 121

Now,

u = -b + - √(b² - 4ac) ÷ 2a

= - 7 + - √121 ÷ 2 × 6

= - 7 + - 11 ÷ 12

= - 7 + 11 / 12 or - 7 - 11 /12

= 4/12 or - 18/12

= 1/3 or - 3/2

Therefore, u = 1/3 or u = - 3/2

sum of zeroes = 1/3 + (- 3/2)

= - b/a

= -(coefficient of x)/ coefficient of x²

= 7/6

product of zeroes = 1/3 × ( - 3/2)

= c/a

= constant term / coefficient of x²

= - 3/6 = - 1/2