1. Show that the rods of the equation $$ (r+1) x^{2}+4 x+1=r $$ are real for all real values of r.

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To show that the rods of the equation $$ (r+1) x^{2}+4 x+1=r $$ are real for all real values of $$ r $$, we need to analyze the quadratic equation in the standard form $$ ax^2 + bx + c = 0 $$.

First, we can rearrange the given equation:

$$
(r+1)x^2 + 4x + (1 - r) = 0
$$

Here, we identify the coefficients:
- $$ a = r + 1 $$
- $$ b = 4 $$
- $$ c = 1 - r $$

For the roots of a quadratic equation to be real, the discriminant must be non-negative. The discriminant $$ D $$ is given by:

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

Substituting the coefficients into the discriminant formula:

$$
D = 4^2 - 4(r + 1)(1 - r)
$$

Now, let's simplify this expression step by step:

1. Calculate $$ 4^2 $$:
   $$
   4^2 = 16
   $$

2. Calculate $$ 4(r + 1)(1 - r) $$:
   $$
   4(r + 1)(1 - r) = 4[(r + 1)(1 - r)] = 4[r - r^2 + 1 - r] = 4[1 - r^2]
   $$

3. Substitute back into the discriminant:
   $$
   D = 16 - 4(1 - r^2) = 16 - 4 + 4r^2 = 12 + 4r^2
   $$

Now, we have:

$$
D = 12 + 4r^2
$$

Since $$ 4r^2 $$ is always non-negative for all real $$ r $$, and $$ 12 $$ is a positive constant, we can conclude that:

$$
D \geq 12 > 0
$$

Thus, the discriminant $$ D $$ is always positive for all real values of $$ r $$. This means that the quadratic equation has two distinct real roots for any real $$ r $$.

Therefore, we have shown that the rods of the equation $$ (r+1)x^2 + 4x + 1 = r $$ are real for all real values of $$ r $$.
For all real values of $$ r $$, the equation $$ (r+1)x^2 + 4x + 1 = r $$ has two real roots because the discriminant $$ D = 12 + 4r^2 $$ is always positive.

1. Show that the rods of the equation \( (r+1) x^{2}+4 x+1=r \) are real for all real values of r.

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