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The quadratic expression $(2 x+1)^{2}-p x+q \neq 0$ for any real $x,$ if
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Verified Answer
The correct answer is:
$p^{2}-8 p-16 q < 0$
Given equation is $(2 x+1)^{2}-p x+q \neq 0$
$$
\begin{array}{lr}
\Rightarrow & 4 x^{2}+4 x+1-p x+q \neq 0 \\
\Rightarrow & 4 x^{2}+(4-p) x+(1+q) \neq 0
\end{array}
$$
Now. D $ < $ 0
$\Rightarrow \quad(4-p)^{2}-4(4)(1+q) < 0$
$\Rightarrow \quad 16-8 p+p^{2}-16-16 q < 0$
$$
\Rightarrow \quad p^{2}-8 p-16 q < 0
$$
$$
\begin{array}{lr}
\Rightarrow & 4 x^{2}+4 x+1-p x+q \neq 0 \\
\Rightarrow & 4 x^{2}+(4-p) x+(1+q) \neq 0
\end{array}
$$
Now. D $ < $ 0
$\Rightarrow \quad(4-p)^{2}-4(4)(1+q) < 0$
$\Rightarrow \quad 16-8 p+p^{2}-16-16 q < 0$
$$
\Rightarrow \quad p^{2}-8 p-16 q < 0
$$
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