If a root of the equation ax^2 + bx + c = 0 i | vedclass

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Question

(A) Let $\alpha$ be a root of the first equation,then $\frac{1}{\alpha}$ is a root of the second equation.For the first equation: $a\alpha^2 + b\alpha

Vedclass · Accepted Answer

$(cc' - aa')^2 = (ba' - cb')(ab' - bc')$

Vedclass · Answer

(A) Let $\alpha$ be a root of the first equation,then $\frac{1}{\alpha}$ is a root of the second equation.For the first equation: $a\alpha^2 + b\alpha + c = 0$.For the second equation: $a'(\frac{1}{\alpha})^2 + b'(\frac{1}{\alpha}) + c' = 0$,which simplifies to $c'\alpha^2 + b'\alpha + a' = 0$.Using the cross-multiplication method for the system of equations $a\alpha^2 + b\alpha + c = 0$ and $c'\alpha^2 + b'\alpha + a' = 0$:$\frac{\alpha^2}{ba' - b'c} = \frac{\alpha}{cc' - aa'} = \frac{1}{ab' - bc'}$.From the second and third terms: $\alpha = \frac{cc' - aa'}{ab' - bc'}$.From the first and third terms: $\alpha^2 = \frac{ba' - b'c}{ab' - bc'}$.Equating $\alpha^2 = (\alpha)^2$,we get: $\frac{ba' - b'c}{ab' - bc'} = \left(\frac{cc' - aa'}{ab' - bc'}\right)^2$.$(cc' - aa')^2 = (ba' - b'c)(ab' - bc')$.

If a root of the equation $ax^2 + bx + c = 0$ is the reciprocal of a root of the equation $a'x^2 + b'x + c' = 0$,then:

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