MCQ
The value of $\lambda$ for which $\left(x^2+4 x+\lambda\right)$ is a perfect square, is
- A$1$
- B$16$
- ✓$4$
- D$9$
✓
Answer
Correct option: C.
$4$
For a quadratic equation to be a perfect square
$D=0$
$b^2-4 a c=0$
$(4)^2-4(1)(\lambda)=0$
$16-4 \lambda=0$
$16=4 \lambda$
$\lambda=4$
$D=0$
$b^2-4 a c=0$
$(4)^2-4(1)(\lambda)=0$
$16-4 \lambda=0$
$16=4 \lambda$
$\lambda=4$
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