How do you solve \frac { 2x - 1} { x + 2} - \frac { 1} { x } \leq 1?

1 Answer
Dec 23, 2017

x:(-2,2-sqrt(6)]uu(0,2+sqrt(6)]

Explanation:

(2x-1)/(x+2)-1/x<=1

x/x*(2x-1)/(x+2)-(x+2)/(x+2)*1/x<=1

((2x^2-x)-(x+2))/(x(x+2))<=1

(2x^2-2x-2)/(x(x+2))<=1

At this point, we must look at three cases. One where x is positive, and one where x is between -2 and 0, and one where x is less than -3. This is because the denominator reads x(x+2). As you may know, multiplying an inequality by a negative value changes the direction of the inequality so we must take these cases.

We also know that x!=0 and x!=2 because then we would be dividing by 0.

Case 1: Assume x>0. Then x(x+2)>0. Thus, no sign changes.

2x^2-2x-2<=x^2+2x

x^2-4x-2<=0

From this equation, we need to determine when it is less than 0. To do so, we can figure out where the quadratic intersects the x-axis. Since it opens up (the coefficient of the x^2 term is greater than 0), we know that the x values between the zeroes are where the quadratic is negative.

x=(-(-4)+-sqrt((-4)^2-4*1*(-2)))/(2*1)

x=(4+-sqrt(24))/2

x=2+-sqrt(6)

That means the quadratic is negative for x-values on the interval (2-sqrt(6)),(2+sqrt(6)).

We only take the positive case because 2-sqrt(6)<0, but we defined x>0 for this case. Thus, we have found one interval:

x:(0,2+sqrt(6)]

Case 2; x:(-2,0). If this is true, then x<0 but x+2>0. Thus, the term x(x+2) is multiplying a negative and a positive, so x(x+2)<0. That means we have to flip the direction of the inequality when we multiply by x(x+2).

2x^2-2x-2>=x^2+2x

x^2-4x-2>=0

Based on the math above, we know the quadratic is positive or zero when x<=2-sqrt(6) or x>=2+sqrt(6). Since we know x:(-2,0), we have found another interval.

x:(-2,2-sqrt(6)].

Finally, case 3: x<-2. In this case, both x and x+2 are negative, so their product is positive. That means our algebra and solutions will look very similar to Case 1.

x:(2-sqrt(6)),(2+sqrt(6))

Since the interval of solutions does not occur in our case of x<-2, there are no solutions in this case.