First slide

Theory of equations

Question

If the equation $x^{2} - 3 px + 2 q = 0 and x^{2} - 3 ax + 2 b = 0$ have a common roots and the other roots of the second equation
is the reciprocal of the other roots of the first, then $(2 q - 2 b)^{2}$

Moderate

A

$36 pa (q - b)^{2}$
B

$18 pa (q - b)^{2}$
C

$36 bq (p - a)^{2}$
D

$18 bq (p - a)^{2}$

Solution

According to the question
$α, β$ are roots of equation $x^{2} - 3 px + 2 q = 0$
and $α, \frac{1}{β}$ are roots of equation $x^{2} - 3 ax + 2 b = 0$
Therefore, we have $α + \frac{1}{β} = 3 a, \frac{α}{β} = 2 b$
and $α + \frac{1}{β} = 3 a, \frac{α}{β} = 2 b$
$\begin{aligned} ∴ (2 q - 2 b)^{2} = {(αβ - \frac{α}{β})}^{2} = α^{2} {(β - \frac{1}{β})}^{2} \\ = \frac{α}{β} \cdot βα {[(α + β) - (α + \frac{1}{β})]}^{2} \\ = (2 b) (2 q) [3 p - 3 a]^{2} \\ = 36 bq (p - a)^{2} \end{aligned}$

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