Solve the equation by extracting square roots. When a solution is irrational, list both the exact solution and its approximation rounded to two dedimal places. (If a solution is rabional, enter fat for ts apporimation.) $$ 4 x^{2}=16 $$ smaller value $$ \quad x=\square=\square $$ larger value $$ \quad x=\square $$

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Solve the quadratic equation by following steps:
- step0: Solve using square roots:
  $$4x^{2}=16$$
- step1: Divide both sides:
  $$\frac{4x^{2}}{4}=\frac{16}{4}$$
- step2: Divide the numbers:
  $$x^{2}=4$$
- step3: Simplify the expression:
  $$x=\pm \sqrt{4}$$
- step4: Simplify the expression:
  $$x=\pm 2$$
- step5: Separate into possible cases:
  $$\begin{align}&x=2\&x=-2\end{align}$$
- step6: Rewrite:
  $$x_{1}=-2,x_{2}=2$$
To solve the equation $$ 4x^{2} = 16 $$ by extracting square roots, we can follow these steps:

1. **Isolate $$ x^{2} $$**:
   $$
   x^{2} = \frac{16}{4} = 4
   $$

2. **Take the square root of both sides**:
   $$
   x = \pm \sqrt{4}
   $$

3. **Calculate the square roots**:
   $$
   x = -2 \quad \text{and} \quad x = 2
   $$

Now, we can summarize the solutions:

- Smaller value: $$ x = -2 $$
- Larger value: $$ x = 2 $$

Since both solutions are rational, their approximations are simply "fat" (indicating they are rational).

Thus, the final answers are:
- Smaller value $$ \quad x = -2 = \text{fat} $$
- Larger value $$ \quad x = 2 = \text{fat} $$
Smaller value $$ x = -2 = \text{fat} $$ Larger value $$ x = 2 = \text{fat} $$

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