Question

The number of distinct roots of the equation $(x-5)(x-7)(x+6)(x+4)=504$ is

• A. $2$
• B. $3$
• C. $4$

Answer 1

The correct option is C $4$

Given biquadratic equation: $(x-5)(x-7)(x+6)(x+4)=504$
Here $-5+4=-7+6$
$\therefore$ This equation is of the form $(x-a)(x-b)(x-c)(x-d)=A,$ where $a+b=c+d$

Upon rearranging this equation to the above form we get,
$(x-5)(x+4)(x-7)(x+6)=504;$
$\Rightarrow (x^2-x-20)(x^2-x-42)=504$
Assuming $x^2-x-20=y,$ the equation becomes:
$y(y-22)=504$
$\Rightarrow y^2-22y-504=0$
Using completing the square method,
$\Rightarrow y^2-2\left(11\right)y+121-121-504=0$
$\Rightarrow (y-11{)}^2=625$
$\Rightarrow y-11=\pm 25$
$\Rightarrow y-11=25$ and $y-11=-25$
$\Rightarrow y=36$ and $y=-14$

If $y=36$
$\Rightarrow x^2-x-20=36$
$\Rightarrow x^2-x-56=0$
$\Rightarrow x^2-8x+7x-56=0$
$\Rightarrow (x-8)(x+7)=0\Rightarrow x=-7,8$

If $y=-14$
$\Rightarrow x^2-x-20=-14$
$\Rightarrow x^2-x-6=0$
$\Rightarrow x^2-3x+2x-6=0$
$\Rightarrow (x-3)(x+2)=0$
$\Rightarrow x=-2,3$

$\therefore$ The roots are $-7,-2,3,8$.
Thus, there are $4$ distinct roots of the equation.