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Question: If the equation *x*<sup>2</sup> – 3p*x* + 2q = 0 and *x*<sup>2</sup> – 3a*x* + 2b = 0 have a common ...

If the equation x2 – 3px + 2q = 0 and x2 – 3ax + 2b = 0 have a common roots and the other roots of the second equation is reciprocal of the other roots of the first then –

A

36 pa (q – b)2

B

18 pa (q – b)2

C

36 bq (p – a)2

D

18 bq (p – a)2

Answer

36 bq (p – a)2

Explanation

Solution

a be common roots, b is other roots

a + b = 3p, ab = 2q

and a, 1β\frac{1}{\beta}are roots of equation x2 – 3ax + 2b = 0

a + 1β\frac{1}{\beta}= 3a, αβ\frac{\alpha}{\beta}= 2b

\ (2q – 2b)2 = (ab – αβ\frac{\alpha}{\beta})2 = a 2 (b – 1β\frac{1}{\beta})2

= αβ\frac{\alpha}{\beta}. ba[(α+β)−(α+1β)]2\left\lbrack (\alpha + \beta) - \left( \alpha + \frac{1}{\beta} \right) \right\rbrack^{2}

= (2b) (2q) [3p – 3a]2

= 36 bq (p – a)2