If $\alpha$, $\beta$ are the roots of the equation $a{ x }^{ 2 }+bx+x=0$, then the roots of the equation $\left( a+b+c \right) { x }^{ 2 }-\left( b+2c \right) x+c=0$ are

Since alpha and beta are the roots of ax^2 + bx + c = 0, their sum is alpha + beta = -(b divided by a) and their product is alpha times beta = c divided by a. For the new equation (a + b + c)x^2 - (b + 2c)x + c = 0, the sum of its roots is (b + 2c) divided by (a + b + c) and the product is c divided by (a + b + c). Dividing the numerator and denominator by a, the sum becomes (1 + 1 divided by alpha + 1 divided by beta) and the product becomes 1 divided by (1 + 1 divided by alpha + 1 divided by beta). Let the new roots be y, so 1 divided by y = 1 + 1 divided by x, which solves to y = x divided by (x - 1). Therefore, the roots are alpha divided by (alpha - 1) and beta divided by (beta - 1).

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