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(D) Let the roots of $x^2 - bx + c = 0$ be $\alpha, \beta$ and the roots of $x^2 - cx + b = 0$ be $\alpha', \beta'$.The difference of the roots for th

Vedclass · Accepted Answer

$-4$

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(D) Let the roots of $x^2 - bx + c = 0$ be $\alpha, \beta$ and the roots of $x^2 - cx + b = 0$ be $\alpha', \beta'$.The difference of the roots for the first equation is $|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} = \sqrt{b^2 - 4c}$.The difference of the roots for the second equation is $|\alpha' - \beta'| = \sqrt{(\alpha' + \beta')^2 - 4\alpha'\beta'} = \sqrt{c^2 - 4b}$.Given that the roots differ by the same quantity,we have $\sqrt{b^2 - 4c} = \sqrt{c^2 - 4b}$.Squaring both sides,we get $b^2 - 4c = c^2 - 4b$.Rearranging the terms,$b^2 - c^2 = 4c - 4b$.$(b - c)(b + c) = -4(b - c)$.Assuming $b 
eq c$,we divide by $(b - c)$ to get $b + c = -4$.

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

If the roots of the equations $x^2 - bx + c = 0$ and $x^2 - cx + b = 0$ differ by the same quantity,then $b + c$ is equal to

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