Question

find the zero of the following quadratic polynomial and verify the relationship between zeros and the coefficients.​

Answer 1

Step-by-step explanation:

(1) let 16y²+48=0 => y²= -3 => y = ±i√3

=> zeroes are +i√3 and -i√3

where i=√(-1)

verification:- alfa+ beta= i√3+(-i√3) = 0

-b/a = 0/16 = 0

alfa×beta = i√3 × -i√3= 3

c/a= 48/16 = 3

(2) let x²-9=0 => x =±3

=> zeroes are +3 and -3

verification:- alfa+beta = 3+(-3) = 0

-b/a= 0/1= 0

alfa×beta = 3×(-3) = -9

c/a= -9/1 = -9

(3) let t²-14=0 => t= ±√14

=> zeroes are √14 and -√14

verification:- alfa+ beta = √14+(-√14)= 0

-b/a = 0/1= 0

alfa×beta = -14

c/a= -14/1 = -14

(4) let x²+7x+12= 0=> x= -4 and x= -3

=> zeroes are -4 and -3

verification:- alfa+beta = -4+(-3)= -7

-b/a= -7/1= -7

alfa×beta = -4×-3=12

c/a= 12/1= 12