We know that sum of the roots of an equation ax^2 + bx + c = 0 is -b/2a and product of roots of this equation is c/a
We have the equation (a+1)x^2 + (2a+3)x + (3a+4) = 0
Given that
sum of roots = -1
=> -(2a+3)/(a+1) = -1
=> (2a+3)/(a+1) = 1
=> 2a+3 = a+1
=> a = -2
Now,
Product of roots = (3a+4)/(a+1)
= [3*(-2) + 4]/[-2+1]
= -2/-1
= 2
So, product of roots of this equation will be 2.
ALTERNATE METHOD- Given that sum of roots = -1
=> -(2a+3)/(a+1) = -1
=> (2a+3)/(a+1) = 1
Adding 1 on both sides
=> [(2a+3)/(a+1)] + 1 = 2
=> [(2a+3) + (a+1)] / (a+1) = 2
=> (3a+4)/(a+1) = 2
For this equation, product of roots = (3a+4)/(a+1) = 2
We have the equation (a+1)x^2 + (2a+3)x + (3a+4) = 0
Given that
sum of roots = -1
=> -(2a+3)/(a+1) = -1
=> (2a+3)/(a+1) = 1
=> 2a+3 = a+1
=> a = -2
Now,
Product of roots = (3a+4)/(a+1)
= [3*(-2) + 4]/[-2+1]
= -2/-1
= 2
So, product of roots of this equation will be 2.
ALTERNATE METHOD- Given that sum of roots = -1
=> -(2a+3)/(a+1) = -1
=> (2a+3)/(a+1) = 1
Adding 1 on both sides
=> [(2a+3)/(a+1)] + 1 = 2
=> [(2a+3) + (a+1)] / (a+1) = 2
=> (3a+4)/(a+1) = 2
For this equation, product of roots = (3a+4)/(a+1) = 2
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