If the roots of ax^2 + bx + c = 0 are alpha, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the equation $ax^2 + bx + c = 0$,the discriminant is $D_1 = b^2 - 4ac$. The difference of the roots is given by $(\alpha - \beta)^2 = (\alpha

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For the equation $ax^2 + bx + c = 0$,the discriminant is $D_1 = b^2 - 4ac$. The difference of the roots is given by $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = \left( -\frac{b}{a} \right)^2 - 4\left( \frac{c}{a} \right) = \frac{b^2 - 4ac}{a^2}$.For the equation $Ax^2 + Bx + C = 0$,the roots are $\alpha - k$ and $\beta - k$. The difference of these roots is $(\alpha - k) - (\beta - k) = \alpha - \beta$. Thus,the square of the difference of the roots is $(\alpha - \beta)^2$.Using the coefficients of the second equation,the square of the difference of the roots is $\frac{B^2 - 4AC}{A^2}$.Equating the two expressions for $(\alpha - \beta)^2$,we get $\frac{b^2 - 4ac}{a^2} = \frac{B^2 - 4AC}{A^2}$.Rearranging the terms,we find $\frac{B^2 - 4AC}{b^2 - 4ac} = \frac{A^2}{a^2} = \left( \frac{A}{a} \right)^2$.

If the roots of $ax^2 + bx + c = 0$ are $\alpha, \beta$ and the roots of $Ax^2 + Bx + C = 0$ are $\alpha - k, \beta - k$,then $\frac{B^2 - 4AC}{b^2 - 4ac}$ is equal to

Similar Questions

$P$ can complete a piece of work in $16$ days,which $Q$ can complete in $32$ days. If $P$ and $Q$ work on alternate days,find the relationship between Quantity $I$ (Total days taken if $P$ starts) and Quantity $II$ (Total days taken if $Q$ starts).

If the roots of the equation $x^2 + a^2 = 8x + 6a$ are real,then:

If $\frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{c^{2}+a^{2}}+\frac{m-c^{2}}{a^{2}+b^{2}}=3,$ then the value of $m$ is

Find the combined product of $(a^{4}+b^{4})(a^{2}+b^{2})(a+b)(a-b)$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $(289)^{\frac{1}{2}} x - \sqrt{324} = 203$
$II.$ $(484)^{\frac{1}{2}} y - \sqrt{225} = 183$

Vedclass Test Series

Exam Paper Generator

Online Exam Module