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(C) Given that $\alpha$ and $\beta$ are roots of the quadratic equation $ax^{2}+bx+c=0$.Therefore, $\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \f

Vedclass · Accepted Answer

$\alpha=3\beta$ or $\beta=3\alpha$

Vedclass · Answer

(C) Given that $\alpha$ and $\beta$ are roots of the quadratic equation $ax^{2}+bx+c=0$.Therefore, $\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Given the condition $3b^{2} = 16ac$.Dividing both sides by $a^{2}$, we get $3\left(\frac{b}{a}ight)^{2} = 16\left(\frac{c}{a}ight)$.Substituting the values of sum and product of roots, we get $3(-\alpha-\beta)^{2} = 16\alpha\beta$.$3(\alpha^{2}+\beta^{2}+2\alpha\beta) = 16\alpha\beta$.$3\alpha^{2}+3\beta^{2}+6\alpha\beta = 16\alpha\beta$.$3\alpha^{2}-10\alpha\beta+3\beta^{2} = 0$.Dividing by $\beta^{2}$ (assuming $\beta 
eq 0$), we get $3\left(\frac{\alpha}{\beta}ight)^{2}-10\left(\frac{\alpha}{\beta}ight)+3 = 0$.Let $x = \frac{\alpha}{\beta}$, then $3x^{2}-10x+3 = 0$.$3x^{2}-9x-x+3 = 0 \Rightarrow 3x(x-3)-1(x-3) = 0$.$(3x-1)(x-3) = 0$.So, $x = 3$ or $x = \frac{1}{3}$.This implies $\frac{\alpha}{\beta} = 3$ or $\frac{\alpha}{\beta} = \frac{1}{3}$.Therefore, $\alpha = 3\beta$ or $\beta = 3\alpha$.

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2}+bx+c=0$ and $3b^{2}=16ac$, then:

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