Question

The value of $a$ for which the quadratic equation $3x^2+2a^{2}x+2x+a^2-3a+2=0$ possesses roots of opposite signs lies in

• A. $(-\infty ,1)$
• B. $(-\infty ,0)$
• C. $(1,2)$
• D. $(\frac{3}{2},2)$

Answer 1

Answer: (C)

$(1,2)$

$3x^2+2a^{2}x+2x+a^2-3a+2=0$

$3x^2+2(a^2+1)x+(a^2-3a+2)=0$

Condition for real distinct roots $\Rightarrow D>0$
$b^2-4ac>0$

$4{(a^2+1)}^2-4\times 3\times (a^2-3a+2)>0$

$4(a^4+1+2a^2)-(12a^2-36a+24)>0$

$4a^4+4+8a^2-12a^2+36a-24>0$

$4a^4-4a^2+36a-20>0$

$a^4-a^2+9a-5>0⟶\left(i\right)$

If roots are of opposite signs then product of roots $<0$

$\frac{constant}{coefficientof{x}^2}<0\Rightarrow \frac{a^2-3a+2}{3}<0$

$a^2-3a+2<0$

$a^2-2a-a+2<0$

$a(a-2)-1(a-2)<0\Rightarrow (a-2)(a-1)<0$

$(a-2)>0$ and $(a-1)<0$ or $(a-2)<0$ and $(a-1)>0$

$a>2$ and $a<1$ or $a<2$ and $a>1$

$a<1$ and $a>2\Rightarrow 2<a<1$ which is not possible

$\therefore a>2$ & $a<1$ is rejected

$\therefore a\in (1,2)$

Hence, C is correct.