Question

The set of values of $p$ for which $6$ lies between the roots of the equation $x^2+2(p-3)x+9=0$ is

• A. $(\frac{3}{4},\infty )$
• B. $(-\infty ,-\frac{3}{4})$
• C. $(-\frac{3}{4},\frac{3}{4})$
• D. $[-\frac{3}{4},\frac{3}{4}]$

Answer 1

Answer: (B)

$(-\infty ,-\frac{3}{4})$

Given $x^2+2(p-3)x+9=0$

roots are real

$\Rightarrow$Discriminant $D\ge 0$
$\Rightarrow \left(2\right(p-3){)}^2-4(9)\ge 0$
$\Rightarrow (p-3{)}^2-9\ge 0$
$(p-3-3)(p-3+3)\ge 0$
$p(p-6)\ge 0$
$p\in (-\infty ,0)\bigcup (6,\infty )$

for roots to exists $6$ to lie between roots, $f\left(6\right)<0$
$\Rightarrow 36+2(p-3)6+9<0$
$12p+9<0$
$\Rightarrow p<\frac{-3}{4}$
$\therefore p\in (-\infty ,\frac{-3}{4})$ for roots to exists and to enclose 6 in between them .