# Polynomials

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## Roots

x3 – 18x2 + bx – c = 0 has positive real roots, p, q and z. If geometric mean of the roots is 6, find b.
1. 36
2. -216
3. 108
4. -72

Choice C. 108

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## Detailed Solution

p + q + z = 18
pq + qz + zp = b
pqz = c
According to the question, ${(pqz)^ \frac{1}{3}}$ = 6. Hence, pqz = c = 216.
This is possible only when p = q= z = 6. {Arithmetic mean of p, q and z = 6, GM is also 6. This is possible only when all three numbers are equal.}
So, pq + qz + zp = 36 × 3 = 108 = b.
Hence, b = 108.

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