If - and - be the values of x obtained from the equation m 2 x 2 -x -2mx-3-0 and if m1 and m2 be the two values of m for which - and - are connected by the relation - - - 4 3- find the value of m 1 2 m 2 - m 2 2 m 1-

The correct option is A 68 3 m ^ 2 ( x ^ 2 x ) nbsp; +2mx+3=0 m^2x^2+(2m m^2)x+3=0 ⇒α+β = m^2 2mm^2=m 2m, αβ = 3m^2 Given, nbsp; ...

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Byju's Answer

Standard XII

Mathematics

Sum of Infinite Terms

if α and ...

Question

if $α$ and $β$ be the values of $x$ obtained from the equation $m^{2} (x^{2} - x) + 2 m x + 3 = 0$ and if $m_{1}$ and $m_{2}$ be the two values of $m$ for which $α$ and $β$ are connected by the relation $\frac{α}{β} + \frac{β}{α} = \frac{4}{3}$ , find the value of $\frac{m_{1}^{2}}{m_{2}} + \frac{m_{2}^{2}}{m_{1}}$ .

- \frac{68}{3}

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- \frac{68}{5}

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- \frac{68}{7}

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- \frac{68}{9}

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Solution

The correct option is A $- \frac{68}{3}$
$m^{2} (x^{2} - x) + 2 m x + 3 = 0$
$m^{2} x^{2} + (2 m - m^{2}) x + 3 = 0$
$\Rightarrow α + β = \frac{m^{2} - 2 m}{m^{2}} = \frac{m - 2}{m}, α β = \frac{3}{m^{2}}$
Given, $\frac{α}{β} + \frac{β}{α} = \frac{4}{3}$
$\Rightarrow \frac{α^{2} + β^{2}}{α β} = \frac{4}{3} \Rightarrow \frac{(α + β)^{2} - 2 α β}{α β} = \frac{4}{3}$
$\Rightarrow \frac{(α + β)^{2}}{α β} = 2 + \frac{4}{3} = \frac{10}{3}$
$\Rightarrow \frac{(m - 2)^{2}}{3} = \frac{10}{3} \Rightarrow m^{2} - 4 m - 6 = 0$
$\Rightarrow m_{1} + m_{2} = 4, m_{1} m_{2} = - 6$
$∴ \frac{m_{1}^{2}}{m_{2}} + \frac{m_{2}^{2}}{m_{1}} = \frac{m_{1}^{3} + m_{2}^{3}}{m_{1} m_{2}} = \frac{(m_{1} + m_{2}) (m_{1}^{2} + m_{2}^{2} - m_{1} m_{2})}{m_{1} m_{2}}$
$= \frac{(m_{1} + m_{2}) [(m_{1} + m_{2})^{2} - 3 m_{1} m_{2}]}{m_{1} m_{2}}$
$= \frac{4 (16 + 3 \times 6)}{- 6} = - \frac{68}{3}$

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