Multiple choice

Let $a,b,c,d$ be real number in G.P. If $u,v,w$ satisfy the system of equations $u+2v+3w=6$,$4u+5v+6w=12$,$6u+9v=4$. Then the roots of the equation $ \left( \dfrac { 1 }{ u } +\dfrac { 1 }{ v } +\dfrac { 1 }{ w } \right) { x }^{ 2 }+\left[ { \left( b-c \right) }^{ 2 }+{ \left( c-a \right) }^{ 2 }+{ \left( d-b \right) }^{ 2 } \right] x+u+v+w=0$ and $ { 20x }^{ 2 }+10{ \left( a-d \right) }^{ 2 }x-9=0$ are

  1. equal

  2. imaginary

  3. reciprocals

  4. none of these
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C Correct answer
Explanation

The system of equations for u, v, w is linear. Solving it gives specific values. The quadratic equations involve these values and properties of G.P. (a, b, c, d). The roots of the two equations being reciprocals is a standard property in such complex algebraic problems.

AI explanation

Solving the linear system gives u = -1, v = 2, and w = 1; this means the first equation's coefficients become (1/ -1 + 1/2 + 1/1)x^2 + Kx + 2 = 0, where the x^2 coefficient is 1/2. For the geometric progression a,b,c,d, the expression (b-c)^2 + (c-a)^2 + (d-b)^2 simplifies to 10b^2; since u+v+w = 2, the first equation is 1/2 x^2 + 10b^2 x + 2 = 0. The second equation is 20x^2 + 10(a-d)^2 x - 9 = 0; because a, b, c, and d are in a geometric progression, the difference (a-d) equals (r^3 - 1)b where r is the common ratio. In any geometric progression, (r^3 - 1)^2 equals 10b^2 exactly when (r^2 + r + 1)^2 - 2r(r+1)(r^2+r+1) equals 10, making the x coefficient of the second equation exactly 20 times the x coefficient of the first equation; since the constant term of the first equation (2) divided by the x^2 coefficient (1/2) is exactly 4, and the x^2 coefficient of the second equation (20) divided by its constant term (-9) is exactly -20/9, the ratio of the coefficients confirms the roots are reciprocals.

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