For the equation ax^2 + bx + c = 0,the roots | vedclass

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(C) The discriminant of the first equation is $D_1 = b^2 - 4ac$. The difference of roots is $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta =

Vedclass · Accepted Answer

$\left( \frac{A}{a} \right)^2$

Vedclass · Answer

(C) The discriminant of the first equation is $D_1 = b^2 - 4ac$. The difference of roots is $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = \frac{b^2}{a^2} - \frac{4c}{a} = \frac{b^2 - 4ac}{a^2}$.For the second equation,the roots are $\alpha - k$ and $\beta - k$. The difference of roots is $(\alpha - k - (\beta - k))^2 = (\alpha - \beta)^2 = \frac{b^2 - 4ac}{a^2}$.Also,for the second equation,the difference of roots squared is $\frac{B^2 - 4AC}{A^2}$.Equating the two,we get $\frac{b^2 - 4ac}{a^2} = \frac{B^2 - 4AC}{A^2}$.Therefore,$\frac{B^2 - 4AC}{b^2 - 4ac} = \frac{A^2}{a^2} = \left( \frac{A}{a} \right)^2$.

Column-$I$	Column-$II$
$A$. $\alpha+\beta+\gamma$	$(1)$ $-\frac{43}{4}$
$B$. $\alpha^2+\beta^2+\gamma^2$	$(2)$ $-\frac{7}{8}$
$C$. $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$	$(3)$ $86$
$D$. $\frac{\alpha}{\beta\gamma}+\frac{\beta}{\gamma\alpha}+\frac{\gamma}{\alpha\beta}$	$(4)$ $0$
	$(5)$ $10$

For the equation $ax^2 + bx + c = 0$,the roots are $\alpha, \beta$,and for the equation $Ax^2 + Bx + C = 0$,the roots are $\alpha - k, \beta - k$. Then $\frac{B^2 - 4AC}{b^2 - 4ac} = \dots$

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