Question

If $\alpha ,\beta and\gamma$ are the real roots of the equation $x^3+5x^2+9x-6=0$, then the value of ${\alpha }^2+{\beta }^2+{\gamma }^2$ is

• A. 07
• B. 7
• C. 7.0
• D. 7.00

Answer 1

Answer: (A), (B), (C), (D)

For a cubic equation $a{x}^3+b{x}^2+cx+d=0,a\ne 0$, with roots $\alpha ,\beta ,\gamma$. Relation between roots and coefficients is given by
$\alpha +\beta +\gamma =\frac{-b}{a},$
$\alpha .\beta +\beta .\gamma +\gamma .\alpha =\frac{c}{a},$
$\alpha .\beta .\gamma =\frac{-d}{a}$

Comparing the given equation $x^3+5x^2+9x-6=0$ with general form, we get $a=1,b=5,c=9,d=-6$

Since
${(\alpha +\beta +\gamma )}^2={\alpha }^2+{\beta }^2+{\gamma }^2+2(\alpha \beta +\beta \gamma +\gamma \alpha )$
$\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2={(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )$
where
$\alpha +\beta +\gamma =-5$ and
$\alpha \beta +\beta \gamma +\gamma \alpha =9$

So, ${\alpha }^2+{\beta }^2+{\gamma }^2$
= ${(-5)}^2-2\times 9$= 25 - 18 = 7