Solve 4(n-1)^2-4(n+1)geq0

$$4 ( n - 1 ) ^ { 2 } - 4 ( n + 1 ) \geq 0$$

Solve for n

$$n\in (-\infty,0]\cup [3,\infty)$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-1\right)^{2}.

$$4\left(n^{2}-2n+1\right)-4\left(n+1\right)\geq 0$$

Use the distributive property to multiply 4 by n^{2}-2n+1.

$$4n^{2}-8n+4-4\left(n+1\right)\geq 0$$

Use the distributive property to multiply -4 by n+1.

$$4n^{2}-8n+4-4n-4\geq 0$$

Combine -8n and -4n to get -12n.

$$4n^{2}-12n+4-4\geq 0$$

Subtract 4 from 4 to get 0.

$$4n^{2}-12n\geq 0$$

Factor out n.

$$4n\left(n-3\right)\geq 0$$

For the product to be ≥0, n and n-3 have to be both ≤0 or both ≥0. Consider the case when n and n-3 are both ≤0.

$$n\leq 0 n-3\leq 0$$

The solution satisfying both inequalities is n\leq 0.

$$n\leq 0$$

Consider the case when n and n-3 are both ≥0.

$$n-3\geq 0 n\geq 0$$

The solution satisfying both inequalities is n\geq 3.

$$n\geq 3$$

The final solution is the union of the obtained solutions.

$$n\leq 0\text{; }n\geq 3$$