Q19 / 82 · NTA Abhyas (JEE Main) 2020 · Mathematics · Quadratic Equation

If $α$ and $β$ are the roots of the equation $8 x^{2} - 3 x + 27 = 0,$ then the value of ${(\frac{α^{2}}{β})}^{\frac{1}{3}} + {(\frac{β^{2}}{α})}^{\frac{1}{3}}$ is

A
$\frac{1}{3}$
Correct answer
B
$\frac{1}{4}$
C
$\frac{1}{5}$
D
$\frac{1}{6}$

Solution

α + β = \frac{3}{8}, α β = \frac{27}{8}

{(\frac{α^{2}}{β})}^{\frac{1}{3}} + {(\frac{β^{2}}{α})}^{\frac{1}{3}} = \frac{{(α^{3})}^{\frac{1}{3}} + {(β^{3})}^{\frac{1}{3}}}{{(α β)}^{\frac{1}{3}}}

\frac{α + β}{{(α β)}^{\frac{1}{3}}} = \frac{\frac{3}{8}}{{(\frac{27}{8})}^{\frac{1}{3}}} = \frac{\frac{3}{8}}{\frac{3}{2}} = \frac{1}{4}

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

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