Q52 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics

$\alpha, \beta$ are the real roots of the equation $12 x^{1 / 3}-25 x^{1 / 6}+12=0$. If $\alpha\gt\beta$, then $\sqrt[6]{\frac{\alpha}{\beta}}=$

  • A
    $\frac{3}{2}$
  • B
    $\frac{4}{3}$
  • C
    $\frac{9}{8}$
  • Correct answer
    D
    $\frac{16}{9}$

Solution

$\begin{aligned} & 12 x^{\frac{1}{3}}-25 x^{\frac{1}{6}}+12=0 . \text { Let } x^{1 / 6}=t \\ & 12 t^2-25 t+12=0, \text { On solving, } t=\frac{4}{3}, \frac{3}{4} \\ & \alpha^{1 / 6}=\frac{4}{3} \beta^{1 / 6}=\frac{3}{4} \Rightarrow \alpha=\left(\frac{4}{3}\right)^6, \beta=\left(\frac{3}{4}\right)^6 \\ & \therefore\left(\frac{\alpha}{\beta}\right)^{1 / 6}=\frac{16}{9} .\end{aligned}$

Hence, the correct answer is option: 'D'.

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