Question

Find the zeroes of the polynomial x2-x-72 and verify the relationship between the zeroes and the coefficient

Answer 1

Answer:

Step-by-step explanation:

p(x)= x^2-x-72=0

=x^2-9x+8x-72=0

=x(x-9)+8(x-9)=0

=(x-9)+(x+8)=0

x-9=0. x+8=0

x=9. x=-8

On comparing ax^2+bx+c=0

We get,

a=1,b=-1 c=-72

Product of zeroes

Alpha×beeta=c/a

9×-8= -72/1

-72=-72

Sum of zeroes

Alpha+beeta=-b/a

9+(-8)=-(-1/1)

9-8=1/1

1=1

Hence,relationship is verified.

Answer 2

Answer:

so A.T.Q

x²-x-72

x²-9x+8x-72

x(x-9)+8(x-9)

(x-9)(x+8)

x=9:x=(-8)

let a =9 and ß=(-8)

so aß=c/a

9(-8)=-72

a+ß=-b/a

9-8=-(-1)

1=1

hence 9 and -8 verify the relation between the zeroes and the coefficients.