Question

find the zeros of polynomial 2x^2-9 and vertify the relationship between the zeros and the coefficient of the polynomial ​

Answer 1

Step-by-step explanation:

first of all split the middle term and it will be in this form

2x^2+6x-x-3

take common

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

(2x-1)(x+3)

x=1/2 , -3

verification

let alpha and beta be the zeroes of the polynomial

alpha=1/2 and beta=-3

alpha+beta=-b/a

1/2+(-3)=-5/2

-6+1/2=-5/2

-5/2=RHS

and aplha×beta=c/a

1/2×-3=-3/2

-3/2= RHS

hence, verified

Answer 2

Answer:

Step-by-step explanation:

Since this is quadratic polynomial there are only two roots or zeroes

Let m and n be the roots / zeroes

=>2x^2-9

=>($\sqrt{2}$x)^2-(3)^2

=>($\sqrt{2}$x-3)($\sqrt{2}$x+3)

so therefore the zeroes of p(x)

=>m=3/$\sqrt{2}$ or n=-3/$\sqrt{2}$

m=3/$\sqrt{2}$ and n= - 3/$\sqrt{2}$

in this eq:-

a=2,b=0,c=-9

m+n=-b/a

3/$\sqrt{2}$ +(-3/$\sqrt{2}$ )=-0/2

0=0

LHS =RHS  

hence verified

mn=c/a

3/$\sqrt{2}$ (-3/$\sqrt{2}$ )=-9/2

-9/2=-9/2

LHS=RHS

hence verified