Question

The values of $a$ for which the quadratic equation $3x^2+2(a^2+1)x+(a^2-3a+2)=0$ has roots of opposite sign, is

• A. $a\in (1,2)$
• B. $a\in (2,\infty )$
• C. $a\in (1,\infty )$
• D. $a\in (2,3)$

Answer 1

The correct option is A $a\in (1,2)$

Given equation is $3x^2+2(a^2+1)x+(a^2-3a+2)=0$
It is an upward opening parabola.
As the roots are of opposite sign, so
$f\left(0\right)<0\Rightarrow a^2-3a+2<0\Rightarrow (a-1)(a-2)<0\therefore a\in (1,2)$