Question

The number of real integral solution(s) of the equation, $(x+9)(x-3)(x-7)(x+5)=385$ is(are)

• A. 2

Answer 1

The correct option is A 2
The given equation is $(x+9)(x-3)(x-7)(x+5)=385$.
This is a type 2 biquadratic equation.
On rearranging the terms and opening the bracket, we get,
$(x+9)(x-7)(x-3)(x+5)=385$
$\Rightarrow (x+9)(x-7)(x+5)(x-3)=385$
$\Rightarrow (x^2+2x-63)(x^2+2x-15)=385$
Put $x^2+2x=z$
$\Rightarrow (z-63)(z-15)=385$
$\Rightarrow z^2-78z+945=385$
$\Rightarrow z^2-78z+560=0$
$\Rightarrow z^2-70z-8z+560=0$
$\Rightarrow (z-70)(z-8)=0$
$\Rightarrow z=70$ or $8$

$x^2+2x=70$
$\Rightarrow x^2+2x-70=0$
$\Rightarrow x=\frac{-2\pm \sqrt{4+280}}{2}$
$\Rightarrow x=\frac{-2\pm \sqrt{284}}{2}$
$\Rightarrow x=-1+\sqrt{71}\$or\$-1-\sqrt{71}$

$x^2+2x=8$
$\Rightarrow x^2+2x-8=0$
$\Rightarrow x=\frac{-2\pm \sqrt{4+32}}{2}$
$\Rightarrow x=2$ or $-4$

$\therefore x=2,-4,-1+\sqrt{71},-1-\sqrt{71}$
Hence, the number of real integral solutions is$2$ i.e $x=2,-4$