Question

The number of integral values of $k$ for which $x^2-(k-1)x+3=0$ has both roots positive and $x^2+3x+6-k=0$ has both roots negative are

• A. $0$
• B. $2$
• C. $1$
• D. Infinite

Answer 1

The correct option is C $1$

The roots of the equation $x^2-(k-1)x+3=0$ are positive .
The roots are real . Hence the discriminant must be non-negative .
$\Rightarrow (k-1{)}^2-12\ge 0$
$\Rightarrow k\ge 1+\sqrt{12}$ or $k<1-\sqrt{12}$
Since, the roots are positive, the product and the sum of the roots must be positive as well .
Product of the roots $=3$
Sum of roots $=k-1>0$
$k>1$.
Hence, the common values for $k$ are $k\ge 1+\sqrt{12}$
For the second equation,
$\Delta =9-4(6-k)>0$
$\Rightarrow k>\frac{15}{4}$
Since the roots are negative, the sum of the roots must be negative and the product of the roots must be positive .
Sum $=-3$
Product $=6-k>0$
$\Rightarrow k<6$
So, the common set is $\frac{15}{4}<k<6$
Hence, the common values for $k$ from both the equations are
$1+\sqrt{12}\le k<6$
The only integral value $k$ can assume is 5 .