Question

Question 12
If a, ß are the roots of the equation ax^2
+ bx +C =0 then the roots of the equation ax? + bx (x+1)+c(x + 1) = 0 are​

Answer 1

Answer:

Step-by-step explanation:

In the equation, ax² + bx + c = 0

Sum of roots = α + β = -b/a

Product of roots = αβ = c/a.

Now, in the equation ax² + bx(x+1) + c(x+1) = 0

=> ax² + bx² + bx + cx + c = 0

=> x²(a+b) + x(b+c) + c = 0

Sum of roots = - (b+c)/a+b

//Divide numerator and denominator by 'a'

                      = -b/a - c/a / a/a + b/a

                      = -(α + β) - αβ / 1 - (α + β)

                      = -1(αβ + α + β)/[1 - (α + β)]

                      = αβ + α + β / α + β - 1

Product of roots = c/a+b

//Divide numerator and denominator by 'a'

                           = c/a / a/a + b/a

                           = (c/a) / 1 + (b/a)

                           = αβ / 1 - (α + β)

Thus the roots of equation are αβ + α + β / α + β - 1 and αβ / 1 - (α + β).