Question

If ${(x^2+x+3)}^2-(p-3)(x^2+x+3)(x^2+x+2)$ $+(p-4){(x^2+x+2)}^2=0$ has at least one real root, then range of $p$ is

• A. $[-5,\frac{39}{7}]$
• B. $(5,\frac{39}{7}]$
• C. $(5,-\frac{39}{7})$
• D. None of these

Answer 1

The correct option is B $(5,\frac{39}{7}]$

${(x^2+x+3)}^2-(p-3)(x^2+x+3)(x^2+x+2)+(p-4){(x^2+x+2)}^2=0$
${(x^2+x+3)}^2-(x^2+x+3)(x^2+x+2)$

$-(p-4)(x^2+x+3)(x^2+x+2)+(p-4){(x^2+x+2)}^2=0$
Factorizing the above equation,
$((x^2+x+3)-(x^2+x+2))((x^2+x+3)-(p-4\left)(x^2+x+2)\right)=0$
$\left(\right(5-p)x^2+(5-p)x+(11-2p\left)\right)=0$
For at least one root, $(5-p{)}^2>4(11-2p\left)\right(5-p)$
Hence, $(5,\frac{39}{7}]$