If the sum of any two roots of the equation x | vedclass

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(A) Let the roots of the equation $x^3+p x^2+q x+r=0$ be $\alpha, \beta, \gamma$. Given that the sum of two roots is zero,let $\alpha+\beta=0$. From V

Vedclass · Accepted Answer

$r=p q$

Vedclass · Answer

(A) Let the roots of the equation $x^3+p x^2+q x+r=0$ be $\alpha, \beta, \gamma$. Given that the sum of two roots is zero,let $\alpha+\beta=0$. From Vieta's formulas: $(i) \alpha+\beta+\gamma = -p$ $(ii) \alpha\beta+\beta\gamma+\gamma\alpha = q$ $(iii) \alpha\beta\gamma = -r$ Since $\alpha+\beta=0$,substituting this into $(i)$ gives $0+\gamma = -p$,so $\gamma = -p$. Substituting $\gamma = -p$ into $(iii)$,we get $\alpha\beta(-p) = -r$,which implies $\alpha\beta = \frac{r}{p}$. Now,substitute $\alpha+\beta=0$ and $\alpha\beta = \frac{r}{p}$ into $(ii)$: $\alpha\beta + \gamma(\alpha+\beta) = q$ $\frac{r}{p} + (-p)(0) = q$ $\frac{r}{p} = q$ $r = pq$ Thus,the correct option is $A$.

Molecule type	$\frac{C_P}{C_V}$
$A$. Monoatomic	$I$. $\frac{7}{5}$
$B$. Diatomic rigid molecules	$II$. $\frac{9}{7}$
$C$. Diatomic non-rigid molecules	$III$. $\frac{4}{3}$
$D$. Triatomic rigid molecules	$IV$. $\frac{5}{3}$

If the sum of any two roots of the equation $x^3+p x^2+q x+r=0$ is zero,then

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