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

The correct option is C $p^2-8p-16q<0$

Given: quadratic expression $(2x+1{)}^2-px+q\ne 0$ for any real $x$

To find the criteria to satisfy above condition.

Sol: The given quadratic expression can be simplified as

$(2x+1{)}^2-px+q\ne 0⟹4x^2+4x+1-px+q\ne 0⟹4x^2+(4-p)x+(q+1)\ne 0$

This is of the form $a{x}^2+bx+c=0$, therefore $a=4,b=(4-p),c=(q+1)$

The given criteria can be satisfied if

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