Question

Find the sum and the products of the zeroes of t^2-16

Answer 1

Let the equation be = ax² + bx + c = 0

Sum of zeroes = -b

Product of zeroes = c

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For this question, sum = 0

Product = -16

Answer 2

Answer:

t^2-16

=t^2-4^2

=(t+4)(t-4)

so,

t+4=0 or t-4=0

=>t=(-4). =>t=4

the zeroes are 4 and (-4)

now,

sum of the zeroes=4+(-4)

=0

=(-0)/1

=(-b/a)

product of the zeroes=4×(-4)

=(-16)

=c/a

Step-by-step explanation:

hope this help u✌️...........