Question

The quadratic equation $3x^2+2(a^2+1)x+a^2-3a+2=0$ possesses roots of opposite sign then a lies in:

• A. $(-\infty ,0)$
• B. $(-\infty ,1)$
• C. $(1,2)$
• D. $(4,9)$

Answer 1

The correct option is C $(1,2)$

The quadratic equation $3x^2+2(a^2+1)x+(a^2-3a+2)=0$ will have two roots of opposite sign if it has real roots as product of the roots is negative, i.e.. if
$D>0and\alpha .\beta <0\Rightarrow 4(a^2+1{)}^2-12(a^2-3a+2)>0and\frac{a^2-3a+2}{3}<0\therefore a^2-3a+2<0$,
i.e., if $(a-1)(a-2)<0or1<a<2$