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Let $a,b,c$ be the roots of $$f(x)=x^3+x^2-5x-1$$ What is the value of the following? $$\lfloor a\rfloor+\lfloor b\rfloor+\lfloor c\rfloor$$

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    Why i add my answer?? – 2017-01-14
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    The roots of this function are very complicated. The only way to realize where the roots are, is to substitute numbers, and check where there is a sign change. For example, $f(2)=1$ and $f(1)<0$, so there's a root between $1$ and $2$. Proceed lkewise. – 2017-01-14
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    The significant of this sign$? – 2017-01-14
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    Between two points where the sign changes, a root is present. By the way, the answer below does your job. – 2017-01-14
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    The $ symbols indicate you can use MathJax in between. E.g. `$x=2$` produces $x=2$. – 2017-01-14

2 Answers 2

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If it weren't for the floor functions, you could use Vieta's formulas. Right now, the easiest way it to look at the graph of the function, which shows you that the roots lie between -3 and -2, between -1 and 0, and between 1 and 2. That means that the floors of these roots are -3, -1 and 1, and adding them up gives the result: -3.

enter image description here

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$x^3+x^2-5x-1=0$ is a cubic equation whose roots are $x=a,b,c$

drawing graph of $f(x) =x^3+x^2-5x-1$

$f'(x) = 3x^2+2x-5$ and $f''(x) = 6x+2$

now local maximum, minimum $f'(x)=0$

so $\displaystyle x = \frac{-2\pm \sqrt{4+60}}{2\cdot 3} = \frac{-2 \pm 8}{6} = 1,-\frac{5}{3}$

so $f''(1) = 8>0$ . so $x=1$ is a point of local minimum and $\displaystyle f''(-\frac{5}{3})<0$

is a point of local maximum

https://www.wolframalpha.com/input/?i=graph+of+x%5E3%2Bx%5E2-5x-1

so one root of $f(x) = 0$ lie between $-3$ to $-2$ and one is $-1$ to $0$ and other is $1$ to $2$

so $\lfloor a \rfloor = -3$ and $\lfloor b \rfloor = -1$ and $\lfloor c \rfloor = 1$

so $\lfloor a \rfloor+\lfloor b\rfloor+\lfloor c \rfloor = -3-1+1 = -3$

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    Very very thnx Glorfindel – 2017-01-14