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Let a, b, c, d be real numbers in G. P. If u, v, w, satisfy the system of equations u + 2v + 3w = 6 4u + 5v + 6w = 12 6u + 9v = 4 Then show that the roots of the equations (1/u + 1/v + 1/w)x 2 + [(b-c) 2 + (c – a) 2 + (d – b) 2 ] x + u + v + w = 0 And 20x 2 + 10(a- d) 2 x – 9 = 0 are reciprocals of each other.

Let a, b, c, d be real numbers in G. P. If u, v, w, satisfy  the system of equations
 u + 2v + 3w = 6
4u + 5v + 6w = 12
6u + 9v = 4
Then show that the roots of the equations
(1/u + 1/v + 1/w)x2 + [(b-c)2 + (c – a)2  + (d – b)2] x + u + v + w = 0
And 20x2 + 10(a- d)2 x – 9 = 0 are reciprocals of each other.

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
9 years ago
Hello Student,
Please find the answer to your question
Solving the system of equations, u + 2v + 3w = 6,
4u + 5v + 6w = 12 and 6u + 9v = 4
We get u = - 1/3, v = 2/3, w = 5/3
∴ u + v + w = 2, 1/u + 1/v +1/w = - 9/10
Let r be the common ratio of the G. P., a, b, c, d. Then b = ar, c = ar2, d = ar3.
Then the first equation
(1/u + 1/v + 1/w) x2 + [(b – c)2 + (c – a)2 + (d – b)2 ] x + (u +v +w) = 0
Becomes
\frac{9}{10}x2 + [(ar – ar2)2 + (ar2 – a)2 + (ar3 – ar)2 ] x + 2 = 0
i.e. 9x2 – 10a2 (1 - r)2 [ r2 + (r + 1)2 + r2 (r + 1)2]x – 20 = 0
i.e. 9x2 – 10a2 (1 – r)2(r4 + 2r3 + 3r2 + 2r + 1)x – 20 = 0
i.e. 9x2 – 10a2 (1 – r)2 (1 + r + r2)2 x – 20 = 0,
i.e. 9x2 – 10a2 (1 – r3)2 x -20 = 0 ……. (1)
The second equation is
20x2 + 10(a – ar3)2 x – 9 = 0
i.e., 20x2 + 10a2 (1 – r3)2 x – 9 =0 …….. (2)
Since (2) can be obtained by the substitution x → 1/x, equations (1) and (2) have reciprocal roots.

Thanks
Aditi Chauhan
askIITians Faculty

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