two integers whose sum is 3 and product is -108
Answers
Answered by
14
Answer:
Let 2 integers be a, b
A/Q,
a+b=3
ab=-108 => b=-108/a
So,
a+b=3
=> b = 3-(-108) = 3+108 =111 ans.
ab=-108
a= (-108)/111
a= 0.973 approx ans.
Answered by
1
Let n,m be those no.
Then,
- n+m = 3
- nm = - 108
m= -108/n
n+m = 3
n = 3 + 108/n = (3n+108)/n
or,
nm=-108
=》nm = (-108/3)×3 = -36×3 = -36/(n+m)
but this was just an experiment, yo will not do these..XD
ANSWER:
- equn1 ÷ equn2
- nm ÷ (n+m) = -36
- or make in a²+b+c =0 form
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