Quadratic inequality with parameter

Question

Hi I've got this inequality with parameter $a\in R$

$\frac{x+a}{x}\le x+2$

I've solved it but I've got different results than book. I've done it by dividing it into 2 cases. 1. x<0 2. x>0 and then by multiplying both sides by x. Then i got quadratic inequalities.

What I've got is:

for $a=-\frac{1}{4}$ $K={-\frac{1}{2}\cup (0,\infty)}$

for $a<-\frac{1}{4}$ $K=(0,\infty)$

for $a>-\frac{1}{4}$ $K=\langle \frac{-1-\sqrt{1+4a}}{2};0)\cup \langle \frac{-1+\sqrt{1+4a}}{2};\infty)$

My whole solution + results from the book http://postimg.org/image/env3w7kb5/

What have I done bad?

Answer 1

$$\frac{x+a}{x}\leq x+2$$ $$\frac{x+a}{x}- x-2\leq 0$$ $$\frac{x+a+x(-x-2)}{x}\leq0$$ $$\frac{-x^2-x+a}{x}\leq 0$$ The discriminant of the numerator is $\Delta=4a+1$

• If $a>\frac{-1}{4}$ ($\Delta>0$) then $x_1=\frac{-1+\sqrt{4a+1}}{2}$ and $x_2=\frac{-1-\sqrt{4a+1}}{2}$ are two distinct roots of the numerator.$$\frac{(x-x_1)(x-x_2)}{x}\geq 0 $$ So the solution is $[x_2, 0[\cup[x_1, +\infty[$
• If $a=\frac{-1}{4}$ ($\Delta=0$) then $x_0=\frac{-1}{2}$ is a double root of the numerator $$\frac{\left(x+\displaystyle\frac{1}{2}\right)^2}{x}\geq 0$$ So the solution is $\{\frac{-1}{2}\}\cup]0,+\infty[$
• If $a<\frac{-1}{4}$ ($\Delta<0$) then the numerator is strictly negative so the solution is $]0,+\infty[$