Question

.

Answer 1

Another . ? :D

Answer 2

If alpha and beta are the roots of the equation ax^2+bx+c, then the quadratic equation,
ax62-bx(x-1)+c(x-1)^2 has roots

Answer 3

\[\large{ax^2 - bx(x-1) + c(x-1)^2 = ax^2 - bx^2 + bx + cx^2 - 2cx + c}\]

\[\large{=(a-b+c)x^2 - 2cx + c}\]

Answer 4

Since alpha and beta are roots of ax^2+bx+c then:

\[\large{\alpha \beta = \cfrac{c}{a}}\]

\[\large{\alpha + \beta = \cfrac{-b}{a}}\]

Answer 5

yes okay..

Answer 6

Now let the roots of the new equation be x1 and x2, then:

\[\large{x_1 + x_2 = \cfrac{-(-2c)}{a-b+c} = \cfrac{2c}{a-b+c}}\]

\[\large{x_1 x_2 = \cfrac{c}{a-b+c}}\]

Answer 7

yes

Answer 8

\[\large{\implies x_1 + x_2 = \cfrac{2(\cfrac{c}{a})}{1 - \cfrac{b}{a} + \cfrac{c}{a}}}\]

Answer 9

Now we know the values of c/a and b/a in terms of alpha, beta

Answer 10

nice got it

Answer 11

Thus:

\[\large{x_1+x_2 = \cfrac{2\alpha\beta}{1 + \alpha + \beta + \alpha \beta}}\]

Answer 12

Similarly:

\[\large{x_1 x_2 = \cfrac{\cfrac{c}{a}}{1- \cfrac{b}{a} + \cfrac{c}{a}}}\]

\[\large{\implies x_1 x_2 = \cfrac{\alpha\beta}{1+\alpha+\beta+\alpha \beta}}\]

Answer 13

thanx

Answer 14

Your welcome :)