If - - are the roots of the equation 8 x 2-3 x -27-0 then the value of -2-1-3-2-1-3A- 1-3B- 1-3C- 4 D- 7-2

The correct option is B 1/3 α + β = 3/8 . αβ = 27/8 α + β = 3/8 , αβ = 27/8 ⇒ (α^2/β)^1/3 + ( β^2 /α)^1/3 = α + β/(αβ )^1/3 = 1/4 ...

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Byju's Answer

Standard XII

Mathematics

Discriminant

If α, β are t...

Question

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.

$\frac{1}{3}$

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$\frac{1}{3}$

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$\frac{7}{2}$

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$4$

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Solution

The correct option is B
$\frac{1}{3}$

$α + β = \frac{3}{8}$ . $α β = \frac{27}{8}$

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

$\Rightarrow$ ${(\frac{α^{2}}{β})}^{\frac{1}{3}}$ + ${(\frac{β^{2}}{α})}^{\frac{1}{3}}$ = $\frac{α + β}{(α β)^{\frac{1}{3}}}$ = $\frac{1}{4}$

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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}}

Q. 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 :

Q. Let

- \frac{π}{6} < θ < - \frac{π}{12}

. Suppose

α_{1}

and

β_{1}

are the roots of the equation

x^{2} - 2 x s e c θ + 1 = 0

α_{2}

and

β_{2}

are the roots of the equation

x^{2} + 2 x t a n θ - 1 = 0

. If

α_{1} > β_{1}

and

α_{2} > β_{2}

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equals

Q. If α, β are roots of the equation

4 x^{2} + 3 x + 7 = 0, then 1 / α + 1 / β

is equal to
(a) 7/3
(b) −7/3
(c) 3/7
(d) −3/7

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If α,β are the roots of the equation 8x2−3x+27=0 then the value of (α2β)13+(β2α)13 is.

The correct option is B 13 α+β=38 . αβ=278 α+β=38 , αβ=278 ⇒ (α2β)13 + (β2α)13 = α+β(αβ)13 = 14

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.

The correct option is B
$\frac{1}{3}$

$α + β = \frac{3}{8}$ . $α β = \frac{27}{8}$

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

$\Rightarrow$ ${(\frac{α^{2}}{β})}^{\frac{1}{3}}$ + ${(\frac{β^{2}}{α})}^{\frac{1}{3}}$ = $\frac{α + β}{(α β)^{\frac{1}{3}}}$ = $\frac{1}{4}$