If $α, β$ are the roots of the equation $2 x^{2} - 5 x + 16 = 0$ then the value of ${(\frac{α^{2}}{β})}^{\frac{1}{3}} + {(\frac{β^{2}}{α})}^{\frac{1}{3}}$ is :

\frac{1}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{5}{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{5}{12}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{5}{4}$
$α + β = \frac{5}{2}; α β = 8$
${(\frac{α^{2}}{β})}^{\frac{1}{3}} + {(\frac{β^{2}}{α})}^{\frac{1}{3}} \Rightarrow \frac{α^{\frac{2}{3}}}{β^{\frac{1}{3}}} + \frac{β^{\frac{2}{3}}}{α^{\frac{1}{3}}} = \frac{α + β}{{(α β)}^{\frac{1}{3}}} = \frac{\frac{5}{2}}{{(8)}^{\frac{1}{3}}} = \frac{5}{4}$

Suggest Corrections

Similar questions

If $α, β$ 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.

Q. If

α, 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}}

If $α, β$ 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