Find the set of possible values of 'm' for which x^2 - (m^2 - 5m + 5)x + (2m^2 - 3m - 4) = 0 has roots whose sum and product are both less than 1.

We have the equation x^2 - (m^2 - 5m + 5)x + (2m^2 - 3m - 4) = 0;

Sum of roots = (m^2 - 5m + 5)
Product of roots = (2m^2 - 3m - 4)

Given that both the sum and product of roots are < 1.

So,

m^2 - 5m + 5 < 1
=> m^2 - 5m + 4 < 0
=> m^2 - m - 4m + 4 < 0
=> m(m - 1) - 4(m - 1) < 0
=> (m - 4)(m - 1) < 0

To satisfy this, m should lie in (1, 4)

Also,

2m^2 - 3m - 4 < 1
=> 2m^2 - 3m - 5 < 0
=> 2m^2 + 2m - 5m - 5 < 0
=> 2m(m + 1) - 5(m + 1) < 0
=> (m + 1)(2m - 5) < 0

To satisfy this, m should lie in (-1, 5/2)

To satisfy both, our answer should be the intersection of (1, 4) and (-1, 5/2) which comes out to be (1, 5/2)

If the sum of the roots of the equation (a+1)x^2 + (2a+3)x + (3a+4) = 0 is -1, Find the product of the roots.

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