Floored roots

Calculus Level 2

x 3 + x 2 − 5 x − 1 = 0 \large{x^3+x^2-5x-1=0}

Let α , β , γ \alpha,\beta,\gamma be the roots of the equation above.

Evaluate ⌊ α ⌋ + ⌊ β ⌋ + ⌊ γ ⌋ \left\lfloor \alpha \right\rfloor +\left\lfloor \beta \right\rfloor +\left\lfloor \gamma \right\rfloor .

Image Credit: Wikimedia N.Mori .


The answer is -3.

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Tanishq Varshney by Tanishq Varshney , 23, India

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2 solutions

Let f ( x ) = x 3 + x 2 − 5 x − 1 \text{Let } f(x) = x^3+x^2-5x-1

⇒ f ′ ( x ) = 3 x 2 + 2 x − 5 = ( 3 x + 5 ) ( x − 1 ) \Rightarrow f'(x) = 3x^2+2x-5 = (3x+5)(x-1)

⇒ f ′ ′ ( x ) = 6 x + 2 ⇒ { f ′ ′ ( − 5 3 ) < 0 ⇒ maximum f ′ ′ ( 1 ) > 0 ⇒ minimum \Rightarrow f''(x) = 6x +2 \quad \Rightarrow \begin{cases} f''\left(-\frac{5}{3} \right) & < 0 & \Rightarrow \text{maximum} \\ f''\left(1 \right) & > 0 & \Rightarrow \text{minimum} \end{cases}

Since { f ( − 5 3 ) = 5.481 f ( 1 ) = − 4 the three roots are: { α < − 5 3 − 5 3 < β < 1 γ > 1 \text{Since } \begin{cases} f \left(-\frac{5}{3} \right) & = 5.481 \\ f \left(1 \right) & = - 4 \end{cases} \text{the three roots are: } \begin{cases} \alpha < -\frac{5}{3} \\ -\frac{5}{3} < \beta < 1 \\ \gamma > 1 \end{cases}

Let us check: \text{Let us check:}

{ f ( − 3 ) = − 4 f ( − 2 ) = 5 ⇒ − 3 < α < − 2 ⇒ ⌊ α ⌋ = − 3 { f ( − 1 ) = 4 f ( 0 ) = − 4 ⇒ − 1 < β < 0 ⇒ ⌊ β ⌋ = − 1 { f ( 1 ) = − 4 f ( 2 ) = 1 ⇒ 1 < γ < 2 ⇒ ⌊ γ ⌋ = 1 \begin{cases} f(-3) = -4 \\ f(-2) = 5 \end{cases} \quad \Rightarrow - 3 < \alpha < -2 \quad \Rightarrow \lfloor \alpha \rfloor = -3 \\ \begin{cases} f(-1) = 4 \\ f(0) = - 4 \end{cases} \quad \Rightarrow - 1 < \beta < 0 \quad \Rightarrow \lfloor \beta \rfloor = -1 \\ \begin{cases} f(1) = -4 \\ f(2) = 1 \end{cases} \quad \Rightarrow 1 < \gamma < 2 \quad \Rightarrow \lfloor \gamma \rfloor = 1

⇒ ⌊ α ⌋ + ⌊ β ⌋ + ⌊ γ ⌋ = − 3 − 1 + 1 = − 3 \Rightarrow \lfloor \alpha \rfloor + \lfloor \beta \rfloor + \lfloor \gamma \rfloor = -3 - 1 + 1 = \boxed{-3}

16 Helpful 1 Interesting 0 Brilliant 0 Confused

Moderator note:

Simple standard approach.

Sudeep Salgia
Aug 2, 2015

L e t f ( x ) = x 3 + x 2 − 5 x − 1 Let f(x) = x^3 + x^2 - 5x - 1 . Clearly, f f is a continuous function. Let us evaluate f ( x ) f(x) at certain points. f ( 2 ) = 8 + 4 − 10 − 1 = 1 f(2) = 8+4-10-1 = 1 f ( 1 ) = 1 + 1 − 5 − 1 = − 4 f(1) = 1+1-5-1 = -4 f ( 0 ) = 0 + 0 − 0 − 1 = − 4 f(0) = 0+0-0-1 = -4 f ( − 1 ) = − 1 + 1 + 5 − 1 = 4 f(-1) = -1 +1 + 5 -1 =4 f ( − 2 ) = − 8 + 4 + 10 − 1 = 5 f(-2) = -8 + 4 +10 -1 = 5 f ( − 3 ) = − 27 + 9 + 15 − 1 = − 4 f(-3) = -27 +9 +15 -1 = -4 If f ( a ) . f ( b ) < 0 f(a).f(b) < 0 for some a , b ∈ R , a < b a,b \in \mathbb{ R} , a<b , then using Intermediate value theorem , we can say that, ∃ c ∈ ( a . b ) \exists c \in (a.b) such that f ( c ) = 0 f(c) = 0 , i.e., a root exists.

Using the above idea and applying it on the intervals ( − 3 , − 2 ) , ( − 1 , 0 ) (-3,-2) , (-1,0) and ( 1 , 2 ) (1,2) , we can conclude that there exists at least one root in each of these intervals. Since the intervals are distinct, all these roots have to be distinct. Also since a cubic polynomial has exactly three roots, these are all the possible roots. Now, the intervals are chosen such that the floor of the roots can be easily determined and would be a unique value. Therefore,
⌊ α ⌋ + ⌊ β ⌋ + ⌊ γ ⌋ = − 3 − 1 + 1 = − 3 \lfloor \alpha \rfloor + \lfloor \beta \rfloor + \lfloor \gamma \rfloor = -3 -1 + 1 = \boxed{ -3 } .

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Simple standard approach.

A slightly more interesting question would have been to ask for { a } + { b } + { c } \{ a \} + \{ b\} + \{c \} .

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