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

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 \( r \).