# Polynomials

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## Cubic Equation

x3 – 4x2 + mx – 2 = 0 has 3 positive roots, two of which are p and ${\frac{1}{p}}$. Find m.
1. 5
2. -11
3. 8
4. -2

Choice A. 5

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## Detailed Solution

Let the roots be p, ${\frac{1}{p}}$ and q.
${p * \frac{1}{p} * q = 2}$
So, the 3rd root = q = 2
p + ${\frac{1}{p}}$ + 2 = 4
p + ${\frac{1}{p}}$ = 2 {For any real p |p + ${\frac{1}{p} \geq 2}$| }
p2 – 2p + 1 = 0
(p – 1)2 = 0. Hence, p is 1.
The roots are 1, 1 and 2.
(1 × 1) + (1 × 2) + (1 × 2) = 5 = m

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