Question

If the equation x³ – 3x + 1 = 0 has three real roots a, b, g such that α < β < γ, then find the value of {α} + {β} + {γ}, where {,} = F.P.F

Answer 1

Given f(x) = x³ – 3x + 1

f'(x) = 3x² – 3 = 3(x² – 1)

= 3(x + 1)(x – 1)

Clearly, f(x) incresing in (– ∞, – 1) ∪ (1, ∞) and decreasing in (–1, 1)

Now, f(– 2) = – 8 + 6 + 1 = – 1

f(–1) = –1 + 3 + 1 = 3

f(0) = 1 f(1) = 1 – 3 + 1 = –1

f(2) = 8 – 6 + 1 = 3

Now, f(– 2) f(–1) < 0

Thus, there is a root lies in (–2, –1)

So, [a] = –2

Also, f(0)f(1) < 0

Thus, there is a root lies in (0, 1)

So, [b ] = 0

Again, f(1)f(2) < 0

Thus, there is a root lies in (1, 2)

So, [g ] = 1

Since α, β, γ are the roots of x³ – 3x + 1 = 0

so, α + β + γ = 0

Now, {α} + {β} + {γ}

= α – [α] + β – [β] + γ – [γ]

= (α + β + γ) – ([α] + [β] + [γ])

= 0 –(–2 + 0 + 1) = 1.