Question

The quadratic expression $\left(2x+1\right)^{2}-px+q\ne0$ for any real $x$ if

• A. $p^2 - 16p - 8q < 0$
• B. $p^2 - 8p + 16q < 0$
• C. $p^2 - 8p - 16q < 0$
• D. $p^2 - 16p + 8q < 0$

Answer 1

Answer: (C)

$p^2 - 8p - 16q < 0$

Answer 2

The correct answer is C:$p^2-8p-16q<0$
Given equation is $(2 x+1)^{2}-p x +q \neq 0$
$\Rightarrow 4 x^{2}+4 x+1-p x +q \neq 0$
$\Rightarrow 4 x^{2}+(4-p) x+(1+q) \neq 0$
Now, $D<0$
$\Rightarrow (4-p)^{2}-4(4)(1+q)<0$
$\Rightarrow 16-8 p +p^{2}-16-16 q<0$
$\Rightarrow p^{2}-8 p-16 q<0$