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 demonstrate that the rods of the equation \( (r+1) x^{2}+4 x+1=r \) are real for all real values of \( r \), we can rearrange the equation to the standard quadratic form: \[ (r+1)x^2 + 4x + (1 - r) = 0 \] This is a quadratic equation in \( x \), and for the roots to be real, the discriminant must be non-negative. The discriminant \( D \) for a quadratic equation \( ax^2 + bx + c = 0 \) is given by: \[ D = b^2 - 4ac \] Applying this to our equation, where \( a = r + 1 \), \( b = 4 \), and \( c = 1 - r \): \[ D = 4^2 - 4(r + 1)(1 - r) \] \[ D = 16 - 4(r + 1)(1 - r) \] Expanding \( (r + 1)(1 - r) \): \[ (r + 1)(1 - r) = r - r^2 + 1 - r = 1 - r^2 \] Thus, \[ D = 16 - 4(1 - r^2) = 16 - 4 + 4r^2 = 12 + 4r^2 \] Since \( 4r^2 \) is always non-negative and adding 12 keeps it non-negative, we conclude that \( D \geq 12 \). Therefore, the roots of the quadratic equation are real for all real values of \( r \). Thus, the rods represented by this equation are always real regardless of the value of \( r \).