First slide

Theory of equations

Question

$If A.M. of the roots of a quadratic equation is 8 / 5 and A.M. of their reciprocals is 8 / 7, then the equation is$

Moderate

A

$7 x^{2} - 16 x + 8 = 0$
B

$3 x^{2} - 12 x + 7 = 0$
C

$5 x^{2} - 16 x + 7 = 0$
D

$7 x^{2} - 16 x + 5 = 0$

Solution

$\begin{array}{l} Let the roots be α and β \\ ∴ \frac{α + β}{2} = \frac{8}{5} \Rightarrow α + β = \frac{16}{5} - - - - - - (1) \end{array}$
$and \frac{\frac{1}{α} + \frac{1}{β}}{2} = \frac{8}{7} \Rightarrow \frac{α + β}{2 α β} = \frac{8}{7} \Rightarrow \frac{(16 / 5)}{2 (8 / 7)} = α β$
$\begin{array}{l} \Rightarrow α β = \frac{7}{5} \\ So, equation is x^{2} - (\frac{16}{5}) x + \frac{7}{5} = 0 \end{array}$
$or 5 x^{2} - 16 x + 7 = 0$

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