Question

If \alpha and \beta are the roots of the equation 2x^{2} -5x+16=0, then the value of (\alpha ^2 /\beta )^1/3 +(\beta ^2/\alpha )^1/3 is:
OPTIONS ARE:
1/3
1/4
1/5
5/4

Answer 1

Answer:

1/5

Step-by-step explanation:

please do the answer yourself.

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Answer 2

Answer:

$(\frac{\alpha ^{2} }{\beta } )^\frac{1}{3} + (\frac{\beta^{2} }{\alpha } )^\frac{1}{3} = \frac{5}{4}$

Step-by-step explanation:

If $\alpha$ and $\beta$ are the roots of the $ax^{2} +bx+ c=0$, then we have,

Sum of roots =   $\alpha + \beta = \frac{-b}{a}$ and

Product of roots = $\alpha \beta = \frac{c}{a}$

Given,

$\alpha$ and $\beta$ are the roots of the $2x^{2} -5x+16=0$

Comparing the equation $2x^{2} -5x+16=0$ with $ax^{2} +bx+ c=0$ we get,

a = 2

b = -5

c = 16

substituting the value of 'a','b' and 'c' we get

Sum of roots =   $\alpha + \beta = \frac{5}{2}$ and

Product of roots = $\alpha \beta = \frac{16}{2} = 8$

Required to find,$(\frac{\alpha ^{2} }{\beta } )^\frac{1}{3} + (\frac{\beta^{2} }{\alpha } )^\frac{1}{3}$

Substitute $\alpha = \frac{8}{\beta }, \beta = \frac{8}{\alpha }$ we get

$(\frac{\alpha ^{2} }{\beta } )^\frac{1}{3} + (\frac{\beta^{2} }{\alpha } )^\frac{1}{3}$

= $(\frac{\((\frac{8}{\beta })^{2} }{\beta } )^\frac{1}{3} + (\frac{\((\frac{8}{\\\alpha })^ {2} }{\alpha } )^\frac{1}{3}$

= $(\frac{64}{\beta ^2} *\frac{1}{\beta } )^\frac{1}{3} + (\frac{64}{\\\alpha ^2} *\frac{1}{\\\alpha } )^\frac{1}{3}$

= $(\frac{64}{\beta ^3})^\frac{1}{3} + (\frac{64}{\\\alpha ^3})^\frac{1}{3}$

= $\frac{4}{\beta } + \frac{4}{\alpha }$

LCM = $\alpha \beta$

= $\frac{4\alpha +4\beta }{\alpha \beta } = \frac{4(\alpha +\beta )}{\alpha \beta }$

Substituting the value of $\alpha +\beta$ and $\alpha \beta$

= $\frac{4*\frac{5}{2} }{8}$ = $\frac{10}{8} = \frac{5}{4}$

Then$(\frac{\alpha ^{2} }{\beta } )^\frac{1}{3} + (\frac{\beta^{2} }{\alpha } )^\frac{1}{3} = \frac{5}{4}$