Question

Question 2 (vi)
Are the following statements ‘True’ or False’? Justify your answer.
vi) If all three zeroes of a cubic polynomial $x^3+a{x}^2–bx+c$ are positive, then atleast one of a, b and c is non - negative.

Answer 1

vi) false , let $\alpha \beta and\gamma$ the three zeroes of cubic polynomial $x^3+a{x}^2–bx+c$
then, product of zeroes = $(-1{)}^3\frac{constantterm}{coefficientof{x}^3}$
$\Rightarrow \alpha \beta \gamma =-\frac{(+c)}{1}$
$\Rightarrow \alpha \beta \gamma =-c$
Given that, all three zeroes are positive. So, the product of all three zeroes is also positive
i.e., $\alpha \beta \gamma >0$
$\Rightarrow -c>0$
$\Rightarrow c<0$
Now, sum of the zeroes = $\alpha +\beta +\gamma =(-1)\frac{coefficientof{x}^2}{coefficientof{x}^3}$
$\Rightarrow \alpha +\beta +\gamma =-\frac{\alpha }{1}=-a$
But $\alpha \beta and\gamma$ are all positive
Thus, its sum is also positive
So, $\alpha +\beta +\gamma >0$
$\Rightarrow -a>0$
$\Rightarrow a<0$
and sum of the product of two zeroes at a time =$(-1{)}^2\frac{coefficientofx}{coefficientof{x}^3}=\frac{-b}{1}$
$\Rightarrow \alpha +\beta \gamma +\gamma \alpha =-b$
$\therefore$ $\alpha \beta +\beta \gamma +\alpha \gamma >0$
$\Rightarrow \alpha \beta +\beta \gamma ++\alpha \gamma >0\Rightarrow -b>0$
$b<0$
So, the cubic polynomial $x^3+a{x}^2–bx+c,$ has all three zeroes which are positive only when all constants a, b and c are negative.