Theory of EquationsmediumFree

Theory of Equations: Respectively Roots 2px 2qx 2rx Positive Value

JEE Maths question with a full step-by-step solution.

Question

(\alpha, \beta)

(\beta, \gamma)

(\gamma, \alpha)

are respectively the roots of

x^2 - 2px + 2 = 0

;

x^2 - 2qx + 3 = 0

;

x^2 - 2rx + 6 = 0

where

\alpha, \beta, \gamma

are all positive, then the value of

p + q + r

1

2

5

6

correct

Solution

Step 1: From the three equations:

\alpha\beta = 2

\beta\gamma = 3

\gamma\alpha = 6

. Step 2:

(\alpha\beta)(\beta\gamma)(\gamma\alpha) = (\alpha\beta\gamma)^2 = 36 \implies \alpha\beta\gamma = 6

(all positive). Step 3:

\alpha = \dfrac{6}{\beta\gamma} = 2

\beta = \dfrac{6}{\gamma\alpha} = 1

\gamma = \dfrac{6}{\alpha\beta} = 3

. Step 4:

\alpha + \beta = 2p = 3 \implies p = \dfrac{3}{2}

;

\beta + \gamma = 2q = 4 \implies q = 2

;

\gamma + \alpha = 2r = 5 \implies r = \dfrac{5}{2}

p + q + r = \frac{3}{2} + 2 + \frac{5}{2} = 6

Correct answer: (4)

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