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 can rewrite the equation in standard quadratic form. Rearranging the terms: \[ (r + 1)x^2 + 4x + (1 - r) = 0. \] This is a quadratic equation \( Ax^2 + Bx + C = 0 \) where: - \( A = r + 1 \) - \( B = 4 \) - \( C = 1 - r \) For the roots (or "rods") of this quadratic equation to be real, the discriminant \( D \) must be non-negative. The discriminant is given by: \[ D = B^2 - 4AC = 4^2 - 4(r + 1)(1 - r). \] Calculating the discriminant, we have: \[ D = 16 - 4[(r+1)(1-r)] = 16 - 4[(r - r^2 + 1 - r)] = 16 - 4[1 - r^2] = 16 - 4 + 4r^2 = 12 + 4r^2. \] Since \( 4r^2 \) is always non-negative for all real \( r \), \( D \) will always be greater than or equal to 12. Therefore, the discriminant \( D \) is non-negative for all values of \( r \): \[ D \geq 12 > 0. \] This means that the quadratic has two distinct real roots for all real \( r \), proving that the rods are indeed real for any real value of \( r \).