Use the discriminant to tell whether the $$ x^{2}-4 x+1=0 $$

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To determine whether the quadratic equation

$$
x^{2}-4x+1=0
$$

has real solutions, we can use the discriminant. The discriminant $$D$$ of a quadratic equation in the form $$ax^2 + bx + c = 0$$ is given by the formula:

$$
D = b^2 - 4ac
$$

For the given equation, we identify the coefficients:

- $$a = 1$$
- $$b = -4$$
- $$c = 1$$

Now, we can calculate the discriminant:

$$
D = (-4)^2 - 4 \cdot 1 \cdot 1
$$

Let's perform the calculation step by step.

1. Calculate $$b^2$$:
   $$
   (-4)^2 = 16
   $$

2. Calculate $$4ac$$:
   $$
   4 \cdot 1 \cdot 1 = 4
   $$

3. Now, substitute these values into the discriminant formula:
   $$
   D = 16 - 4 = 12
   $$

Now that we have the discriminant $$D = 12$$, we can interpret the result:

- If $$D > 0$$, the equation has two distinct real solutions.
- If $$D = 0$$, the equation has exactly one real solution (a repeated root).
- If $$D < 0$$, the equation has no real solutions (the solutions are complex).

Since $$D = 12 > 0$$, we conclude that the equation

$$
x^{2}-4x+1=0
$$

has two distinct real solutions.
The equation $$x^{2} - 4x + 1 = 0$$ has two distinct real solutions because the discriminant $$D = 12$$ is greater than zero.

Use the discriminant to tell whether the \[ x^{2}-4 x+1=0 \]

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