Write the value of λ for which
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 $\lambda=4$

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