Question

Solve the equation
$(x+2)(x+3)(x+8)(x+12)={4x}^2........\left(1\right)$
find the difference between the rational roots .

Answer 1

Since $(-2)(-12)=(-3)(-8)$ we can

write equation (1) as

$(x+2)(x+12)(x+3)(x+8)={4x}^2$

$\Rightarrow (x^2+14x+24)(x^2+11x+24)=4x^2$ ...(2)

Now $x=0$ is not a root of (1)

Dividing by $x^2$ in both sides of (2), we get

$\Rightarrow (x+\frac{24}{x}+14)(x+\frac{24}{x}+11)=4$ ...(3)

Putting $x+\frac{24}{x}=y,$ we get

$(y+14)(y+11)=4$

$\Rightarrow y^2=25y+150=0$

$\Rightarrow y=-15$ or $y=-10$

Thus the ariginal equation is equivlent to $x+\frac{24}{x}=-15$ or $x+\frac{24}{x}=-10$

$\Rightarrow x^2+15x+24=0$ or $x^2+10x+24=0$

$\Rightarrow x=\frac{-15\pm \sqrt{129}}{2}$ or $x=-6,-4$

$\therefore$ Difference between sational roots is $-4-(-6)=2.$