How to solve 4p ^ 2 <= 1

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

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$$4p^{2}\le1$$

$p\in \begin{bmatrix}-\frac{1}{2},\frac{1}{2}\end{bmatrix}$

Divide both sides by $4$. Since $4$ is positive, the inequality direction remains the same.

$$p^{2}\leq \frac{1}{4}$$

Calculate the square root of $\frac{1}{4}$ and get $\frac{1}{2}$. Rewrite $\frac{1}{4}$ as $\left(\frac{1}{2}\right)^{2}$.

$$p^{2}\leq \left(\frac{1}{2}\right)^{2}$$

Inequality holds for $|p|\leq \frac{1}{2}$.

$$|p|\leq \frac{1}{2}$$

Rewrite $|p|\leq \frac{1}{2}$ as $p\in \left[-\frac{1}{2},\frac{1}{2}\right]$.

$$p\in \begin{bmatrix}-\frac{1}{2},\frac{1}{2}\end{bmatrix}$$