Question

If $\alpha ,\beta ,\gamma$ are roots of $x^3+x^2-5x-1=0$ then $\left[\alpha \right]+\left[\beta \right]+\left[\gamma \right]$

• A. $-1$
• B. $2$
• C. $-3$
• D. $4$

Answer 1

The correct option is C $-3$

Let $f\left(x\right)=x^3+x^2-5x-1$

$f\left(0\right)=0+0-0-1=-1<0$

$f\left(1\right)=1+1-5-1=-4<0$

$f\left(2\right)=8+4-10-1=2-1=1>0$

Considering the negative values,

$f(-1)=(-1{)}^3+(-1{)}^2-5(-1)-1$

$=-1+1+5-1=4>0$

$f(-2)=(-2{)}^3+(-2{)}^2-5(-2)-1$

$=-8+4+10-1=5>0$

$f(-3)=(-3{)}^3+(-3{)}^2-5(-3)-1$

$=27+9+15-1$

$=24-27-1$

$=-4<0$

There is a change of sign between $f(-3)$ and $f(-2)$;

$f(-1)$ and $f\left(0\right)$; $f\left(1\right)$ and $f\left(2\right)$

$\therefore -3<\alpha <-2;-1<\beta <0;1<\gamma <2$

$\therefore \left[\alpha \right]+\left[\beta \right]+\left[\gamma \right]=-3-1+1=-3$

$\left(c\right)-3$.