Give a value of \( c \) in the equation \( 9 x^{2}-12 x+c=0 \) so we have a. no real roots b. one real root c. two real rational roots (give fractions as a solution) d. two real integer roots (give whole numbers as a solution)

To determine the values of \( c \) for each case, we can use the discriminant \( D \) from the quadratic formula \( D = b^2 - 4ac \). The discriminant tells us about the nature of the roots of the quadratic equation \( ax^2 + bx + c = 0 \). a. For no real roots, we need \( D < 0 \). In our case, this means that: \[ D = (-12)^2 - 4 \times 9 \times c < 0 \] \[ 144 - 36c < 0 \] \[ 36c > 144 \] \[ c > 4 \] So, any value of \( c \) greater than 4 will ensure no real roots. For example, \( c = 5 \). b. For one real root, we need \( D = 0 \). Thus: \[ D = 144 - 36c = 0 \] \[ 36c = 144 \] \[ c = 4 \] So, \( c = 4 \) gives one real root. c. For two real rational roots, we need \( D > 0 \) and perfect squares: \[ D = 144 - 36c > 0 \] You could let \( D = k^2 \) where \( k \) is an integer. Therefore, we can choose \( c \) such that \( 144 - 36c \) is a perfect square. Let's express it as: \[ 144 - k^2 = 36c \] Choosing \( k = 12 \): \[ 144 - 12^2 = 0 \implies c = 4 \] (only one root) For two rational roots, choose \( k = 6 \): \[ 144 - 6^2 = 108 \Rightarrow 36c = 108 \Rightarrow c = 3 \] Thus, \( c = 3 \) results in two rational roots. The roots would be \( \frac{2}{3} \) and \( \frac{1}{3} \). d. For two real integer roots, we require \( c \) such that the roots of the equation are integers. The roots can be expressed as \( x_1, x_2 \) satisfying: \[ x_1 + x_2 = \frac{-b}{a} = \frac{12}{9} = \frac{4}{3} \] However, to find \( c \), we can factor the equation for integer roots. Let’s assume the roots are \( m \) and \( n \): \[ (x - m)(x - n) = 9x^2 - 12x + c \] The product \( mn \) should equal \( \frac{c}{9} \). If we let the roots be \( 1 \) and \( 2 \): \[ c = 9 \cdot (1 \cdot 2) = 18 \] So, \( c = 18 \) would yield two integer roots: \( 1, 2 \).