Question

The value of $k$ for which the equation
$3x^2+2x(k^2+1)+k^2-3k+2=0$
has roots of opposite signs, lies in the interval

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

Answer 1

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

Given quadratic equation $3x^2+2x(k^2+1)+k^2-3k+2=0$

Now, for the equation $a{x}^2+bx+c=0$ to have roots of opposite signs $a\cdot c<0$

Comparing the equations, we get:
$a=3,c=k^2-3k+2$
$\therefore \text{for}3x^2+2x(k^2+1)+k^2-3k+2=0$ to have roots of opposite signs,
$3(k^2-3k+2)<0\Rightarrow k^2-3k+2<0\Rightarrow (k-1)(k-2)<0$

Using wavy curve method we find the required range of $k$


$\therefore (k-1)(k-2)<0\Rightarrow k\in (1,2)$