Question

For what value/s of k, the quadratic equation $2k{x}^2+6kx+9=0$ has no real roots?

• A. $0<k<2$
• B. $k=0$
• C. $k>2$
• D. $-2<k<0$

Answer 1

The correct option is A $0<k<2$

$2k{x}^2+6kx+9=0$

$a=2k,b=6k,c=9$

$\text{Since the quadratic equation has no real roots},$

$D=b^2-4ac<0$

$(6k{)}^2-4\times 2k\times 9<0$

$36k^2-72k<0$

$k(k-2)<0$

$\text{It is of the form}(x-a)(x-b)<0anda<b,\text{then}$

$a<x<b$

$0<k<2$

Thus, for all values of k in (0,2), the quadratic equation has no real roots

$\text{(a) is correct}$