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 roots of the equation \( (r+1)x^{2} + 4x + (1-r) = 0 \) are real for all real values of \( r \), we need to examine the discriminant of the quadratic equation. The discriminant \( D \) is given by \( D = b^2 - 4ac \). Here, we have: - \( a = r + 1 \) - \( b = 4 \) - \( c = 1 - r \) Now, the discriminant becomes: \[ D = 4^2 - 4(r + 1)(1 - r) = 16 - 4[(r + 1)(1 - r)] \] Calculating the product: \[ (r + 1)(1 - r) = r - r^2 + 1 - r = 1 - r^2 \] Now substituting back into the discriminant: \[ D = 16 - 4(1 - r^2) = 16 - 4 + 4r^2 = 12 + 4r^2 \] The term \( 12 + 4r^2 \) is always greater than or equal to 12, which is a positive number for all real \( r \). Since the discriminant is non-negative (i.e., \( D \geq 0 \)), it follows that the roots of the equation are real for all real values of \( r \). Thus, we have shown that the roots of the equation \( (r+1) x^{2} + 4x + (1-r) = 0 \) are real for all real \( r \).