MCQ
The value of $\lambda$ for which $(\text{x}_2 + 4\text{x} + \lambda)$ is a perfect square, is :
- A$16$
- B$9$
- C$1$
- ✓$4$
✓
Answer
Correct option: D.
$4$
The given quadric equation is $\text{x}_2 + 4\text{x} + \lambda=0$
Then find the value of $k$.
Here, $a = 1, b = 4$ and, $\text{c}=\lambda$
As we know that $D=b^2-4 a c$
Putting the value of $a = 1, b = 4$ and, $\text{c}=\lambda$
$\lambda=(4)^2-4\times1\times\lambda$
$=16-4\lambda$
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 $\lambda=4$
Then find the value of $k$.
Here, $a = 1, b = 4$ and, $\text{c}=\lambda$
As we know that $D=b^2-4 a c$
Putting the value of $a = 1, b = 4$ and, $\text{c}=\lambda$
$\lambda=(4)^2-4\times1\times\lambda$
$=16-4\lambda$
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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