If alpha, beta, gamma are roots of x^3 + x^2 | vedclass

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Question

(B) Let $f(x) = x^3 + x^2 - 5x - 1$.To find the intervals where the roots lie,we examine the values of $f(x)$ at various points:$f(1) = 1 + 1 - 5 - 1

Vedclass · Accepted Answer

$-3$

Vedclass · Answer

(B) Let $f(x) = x^3 + x^2 - 5x - 1$.To find the intervals where the roots lie,we examine the values of $f(x)$ at various points:$f(1) = 1 + 1 - 5 - 1 = -4 $f(2) = 8 + 4 - 10 - 1 = 1 > 0$Since $f(1)  0$,one root $\alpha$ lies in $(1, 2)$. Thus,$[\alpha] = 1$.$f(0) = -1 $f(-1) = -1 + 1 + 5 - 1 = 4 > 0$Since $f(-1) > 0$ and $f(0) $f(-2) = -8 + 4 + 10 - 1 = 5 > 0$$f(-3) = -27 + 9 + 15 - 1 = -4 Since $f(-3)  0$,one root $\gamma$ lies in $(-3, -2)$. Thus,$[\gamma] = -3$.Therefore,$[\alpha] + [\beta] + [\gamma] = 1 + (-1) + (-3) = -3$.

If $\alpha, \beta, \gamma$ are roots of $x^3 + x^2 - 5x - 1 = 0$,then the value of $[\alpha] + [\beta] + [\gamma]$ is (where $[.]$ denotes the greatest integer function):

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