Find the least positive value of k for which the equation

Solution:

The given quadric equation is $x^{2}+k x+4=0$, and roots are real.

Then find the value of $k$.

Here,

$a=1, b=k$ and,$c=4$

As we know that $D=b^{2}-4 a c$

Putting the value of $a=1, b=k$ and, $c=4$

$=(k)^{2}-4 \times 1 \times 4$

$=k^{2}-16$

The given equation will have real and equal roots, if $D=0$

$k^{2}-16=0$

Now factorizing of the above equation

$k^{2}-16=0$

$k^{2}=16$

$k=\sqrt{16}$

$=\pm 4$

Now according to question, the value of k is positive.

Therefore, the value of $k=4$