Question:
Write the value of $\lambda$ for which $x^{2}+4 x+\lambda$ is a perfect square.
Solution:
The given quadric equation is $x^{2}+4 x+\lambda=0$
Then find the value of $k$.
Here, $a=1, b=4$ and,$c=\lambda$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=1, b=4$ and, $c=\lambda$
$=(4)^{2}-4 \times 1 \times \lambda$
$=16-4 \lambda$
The given equation are perfect square, if $D=0$
$14-4 \lambda=0$
$4 \lambda=16$
$\lambda=\frac{16}{4}$
$=4$
Therefore, the value of $\lambda=4$