given that \( 4(y-2)^{2}=k(y-1) \), where k is a constant. (i) Show hat the sum of the roots equals their product. (ii) Find the value for which the difference between the roots is \( 33 / 4 \)

Certainly! Let's tackle the problem step by step. **Given Equation:** \[ 4(y - 2)^2 = k(y - 1) \] where \( k \) is a constant. --- ### Part (i): Show that the Sum of the Roots Equals Their Product **Step 1: Expand and Rearrange the Equation** First, expand the left side of the equation: \[ 4(y - 2)^2 = 4(y^2 - 4y + 4) = 4y^2 - 16y + 16 \] Now, rewrite the equation by bringing all terms to one side: \[ 4y^2 - 16y + 16 = k(y - 1) \\ 4y^2 - 16y + 16 - ky + k = 0 \\ 4y^2 - (16 + k)y + (16 + k) = 0 \] **Step 2: Identify Coefficients** The equation is now in standard quadratic form: \[ a y^2 + b y + c = 0 \] where: - \( a = 4 \) - \( b = -(16 + k) \) - \( c = 16 + k \) **Step 3: Apply Vieta’s Formulas** For a quadratic equation \( a y^2 + b y + c = 0 \): - **Sum of the roots (\( S \))**: \( S = -\frac{b}{a} \) - **Product of the roots (\( P \))**: \( P = \frac{c}{a} \) Applying Vieta's formulas: \[ S = -\frac{-(16 + k)}{4} = \frac{16 + k}{4} \\ P = \frac{16 + k}{4} \] **Conclusion for Part (i):** \[ S = P \] Thus, the sum of the roots equals their product. --- ### Part (ii): Find the Value of \( k \) for Which the Difference Between the Roots Is \( \frac{33}{4} \) **Step 1: Express the Difference of the Roots** For a quadratic equation, the difference between the roots can be found using: \[ (r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1 r_2 = S^2 - 4P \] Given that \( S = P \): \[ (r_1 - r_2)^2 = S^2 - 4S \] We are given that \( |r_1 - r_2| = \frac{33}{4} \), so: \[ (r_1 - r_2)^2 = \left(\frac{33}{4}\right)^2 = \frac{1089}{16} \] **Step 2: Substitute \( S = \frac{16 + k}{4} \) into the Equation** \[ S^2 - 4S = \frac{1089}{16} \\ \left(\frac{16 + k}{4}\right)^2 - 4\left(\frac{16 + k}{4}\right) = \frac{1089}{16} \] Simplify the equation: \[ \frac{(16 + k)^2}{16} - \frac{16 + k}{1} = \frac{1089}{16} \] Multiply through by 16 to eliminate denominators: \[ (16 + k)^2 - 16(16 + k) = 1089 \\ 256 + 32k + k^2 - 256 - 16k = 1089 \\ k^2 + 16k - 1089 = 0 \] **Step 3: Solve the Quadratic Equation for \( k \)** Use the quadratic formula: \[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = 16 \), and \( c = -1089 \): \[ k = \frac{-16 \pm \sqrt{16^2 - 4(1)(-1089)}}{2(1)} \\ k = \frac{-16 \pm \sqrt{256 + 4356}}{2} \\ k = \frac{-16 \pm \sqrt{4612}}{2} \\ k = \frac{-16 \pm 2\sqrt{1153}}{2} \\ k = -8 \pm \sqrt{1153} \] **Conclusion for Part (ii):** \[ k = -8 + \sqrt{1153} \quad \text{or} \quad k = -8 - \sqrt{1153} \] --- ### Final Answer **(i)** The sum and product of the roots are both equal to \( \frac{16 + k}{4} \), hence they are equal. **(ii)** The value of \( k \) that makes the difference between the roots \( \frac{33}{4} \) is: \[ k = -8 \pm \sqrt{1153} \]