Question
Write the value of $\lambda,$ for which $\text{x}^2+4\text{x}+\lambda$ is a perfect square.
✓
Answer
In $\text{x}^2+4\text{x}+\lambda$
$\text{a}=1,\text{b}=4,\text{c}=\lambda$
$\text{x}^2+4\text{x}+\lambda$ will be a perfect square if $\text{x}^2+4\text{x}+\lambda=0$ has equal roots
$\Rightarrow\text{D}=\text{b}^2-4\text{ac}$
$\Rightarrow\text{D}=(4)^2-4\times1\times\lambda$
$\Rightarrow\text{D}=0$
$\Rightarrow16-4\lambda=0$
$\Rightarrow16=4\lambda$
$\Rightarrow\lambda=4$
Hence $\lambda=4$
$\text{a}=1,\text{b}=4,\text{c}=\lambda$
$\text{x}^2+4\text{x}+\lambda$ will be a perfect square if $\text{x}^2+4\text{x}+\lambda=0$ has equal roots
$\Rightarrow\text{D}=\text{b}^2-4\text{ac}$
$\Rightarrow\text{D}=(4)^2-4\times1\times\lambda$
$\Rightarrow\text{D}=0$
$\Rightarrow16-4\lambda=0$
$\Rightarrow16=4\lambda$
$\Rightarrow\lambda=4$
Hence $\lambda=4$
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