Question

For what values of 'p' for which the quadratic equation ${px}^2-4x+p$ has real linear factors ?

• A. $-2\le p<3$
• B. $-2\le p\le 2$
• C. $-2\ge p\le 2$
• D. None of these

Answer 1

Answer: (B)

$-2\le p\le 2$

Given equation is: $p{x}^2-4x+9=0$

For real linear factors, discriminant of quadratic equation has to be greater than or equal to zero.
$D\ge 0$
$\Rightarrow 16-4p^2\ge 0$
$\Rightarrow p^2\le 4$
Thus,

$-2\le p\le 2$