Question

The number of real roots of the equation $(x+1)(x+2)(x+3)(x+4)=120$ is

• A. $0$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Given equation
$(x+1)(x+2)(x+3)(x+4)=120$
$\Rightarrow (x+1)(x+4)(x+2)(x+3)=120\Rightarrow (x^2+5x+4)(x^2+5x+6)=120$

Assuming $x^2+5x=t$
$\Rightarrow (t+4)(t+6)=120\Rightarrow t^2+10t-96=0\Rightarrow (t+16)(t-6)=0\Rightarrow t=6\text{or}t=-16$
Now,
$x^2+5x=-16\Rightarrow x^2+5x+16=0\Rightarrow D=25-64<0$
Non real roots.

$x^2+5x-6=0\Rightarrow (x+6)(x-1)=0\Rightarrow x=-6,1$

Hence, the number of real roots are $2$.