34. $\frac{x^2+4x+4}{2x^2-x-1}>0$

Expert intro: This is a rational inequality. The key is to factor the numerator and denominator, then study the sign on each interval determined by the critical points.

Detailed walkthrough

Factor the expression first:

$\frac{x^2+4x+4}{2x^2-x-1}>0$

$\frac{(x+2)^2}{(2x+1)(x-1)}>0$

1) Identify the critical points

• Numerator: $(x+2)^2=0$ at $x=-2$
• Denominator: $(2x+1)(x-1)=0$ at $x=-\frac12,\ 1$

So the sign changes can only occur at

$x=-2,\ -\frac12,\ 1$

2) Analyze the sign

Because $(x+2)^2\ge 0$ and is positive except at $x=-2$, the sign of the fraction is determined by the denominator, except that the whole fraction is $0$ at $x=-2$.

We need the fraction to be strictly positive, so:

• numerator must be positive, not zero
• denominator must be positive

The denominator $(2x+1)(x-1)$ is positive on:

$(-\infty,-\tfrac12)\cup(1,\infty)$

Now exclude $x=-2$ because it makes the fraction equal to $0$, not greater than $0$.

3) Final solution

$(-\infty,-2)\cup(-2,-\tfrac12)\cup(1,\infty)$

This is the solution set.

💡 Pitfall guide

A common mistake is to include $x=-2$ because it makes the numerator zero. But the inequality is strict: $>0$, so values that make the expression equal to $0$ must be excluded. Also remember to exclude $x=-\frac12$ and $x=1$ because they make the denominator zero.

🔄 Real-world variant

If the inequality were $\ge 0$ instead of $>0$, then $x=-2$ would be included, giving

$(-\infty,-\tfrac12)\cup(1,\infty)\setminus\{-\tfrac12,1\}$

More precisely:

$(-\infty,-\tfrac12)\cup(1,\infty)$ with $x=-2$ included inside the first interval because the value becomes $0$ there.

🔍 Related terms

rational inequality, sign chart, critical points