How many integer solutions does the inequality (x^2 - 16)/(1 + 4x - 5x^2) >= 0 have?

Step-by-step solution

1. 1

   Factor the numerator and denominator. The numerator is a difference of squares: $x^2-16=(x-4)(x+4)$. For the denominator, factor out the sign first: $1+4x-5x^2=-(5x^2-4x-1)=-(5x+1)(x-1)$.

2. 2

   Rewrite with the leading minus sign pulled out. The inequality becomes $\frac{(x-4)(x+4)}{-(5x+1)(x-1)}\ge 0$, i.e. $\frac{(x-4)(x+4)}{(5x+1)(x-1)}\le 0$. Flipping the direction here is the step that most solutions get wrong.

3. 3

   List the critical values and exclusions. The numerator vanishes at $x=\pm4$ (these are allowed, since the inequality is non-strict) and the denominator vanishes at $x=-\frac15$ and $x=1$ (these must be excluded). In increasing order the breakpoints are $-4,\ -\frac15,\ 1,\ 4$.

4. 4

   Build the sign chart. Testing one point in each interval: at $x=-5$ the quotient is positive, on $(-4,-\frac15)$ it is negative, on $(-\frac15,1)$ it is positive, on $(1,4)$ it is negative, and for $x>4$ it is positive. The inequality $\le 0$ therefore holds on $[-4,-\frac15)\cup(1,4]$.

5. 5

   Collect the integers in those intervals. From $[-4,-\frac15)$: $-4,-3,-2,-1$. From $(1,4]$: $2,3,4$. Note $x=1$ and $x=-\frac15$ are excluded and $x=0$ lies in a positive region, so neither is counted.

6. 6

   Count and spot-check. That is $4+3=7$ integers. Checking $x=-1$ in the original: $\frac{1-16}{1-4-5}=\frac{-15}{-8}=1.875\ge 0$; checking $x=2$: $\frac{4-16}{1+8-20}=\frac{-12}{-11}\approx 1.09\ge 0$; checking $x=5$: $\frac{9}{-104}<0$, correctly excluded.