Question

What is the smallest integral value of $k$ such that $2x(kx-4)-x^2+6=0$ has no real roots?

• A. $-1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

$2x(kx-4)-x^2+6=0$

$\therefore 2k{x}^2-8x-x^2+6=0$

$\therefore (2k-1)x^2-8x+6=0$

The condition for a quadratic equation to have no real roots is $Discriminant<0$

$\therefore (-8{)}^2-4(2k-1\left)\right(6)<0$

$\therefore 64-48k+24<0$

$\therefore k>\frac{88}{48}$

$\therefore k>\frac{11}{6}$

So, the smallest integer value of k is $2$

The answer is option (B)