find zeroes for 4u^2+8u
Answers
Answered by
25
Hello !
_________________________
4u² + 8u
4u(u + 2)
u = 0 or u = -2
so zeros are 2 or 0
__________________________
relation between zeros and coefficient :
sum of zeros = -b/a
-2 + 0 = - 8 /4 = -2
LHS = RHS
Product of zeros
(-2)(0) = c/a = 0/4 = 0
LHS = RHS
_________________________
4u² + 8u
4u(u + 2)
u = 0 or u = -2
so zeros are 2 or 0
__________________________
relation between zeros and coefficient :
sum of zeros = -b/a
-2 + 0 = - 8 /4 = -2
LHS = RHS
Product of zeros
(-2)(0) = c/a = 0/4 = 0
LHS = RHS
Poornadhithya:
can u please find the relation ship b/w zeroes and co-efficients too. please
Yes sure :)
Answered by
27
heya...
solution is in the pic
hope it helps u
solution is in the pic
hope it helps u
Attachments:
hope it helps u dear
do mark it as brainliest if it helps
Nice answer dear :)
^-^
thnks dear
Similar questions