Question

The quadratic expression $(2X+1{)}^2-pX+q\ne 0$ 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)

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

Given equation:

$(2X+1{)}^2-pX+q\ne 0$
$\Rightarrow 4X^2+4X+1-pX+q\ne 0$
$\Rightarrow 4X^2+(4-p)X+(1+q)\ne 0$

Now,

$D<0$
$\Rightarrow (4-p{)}^2-4(4\left)\right(1+q)<0$
$\Rightarrow 16-8p+p^2-16-16q<0$
$\Rightarrow p^2-8p-16q<0$