ou are given that \( x=\left(2 x^{2}-x+8\right) y \) Show, by the method of complet- ing the square that \( \frac{(x-1+y)^{2}}{4 y}=\frac{1+2 y-63 y^{2}}{16 y^{2}} \) i) Find the values of \( y \) for which \( 1+2 y-63 y^{2} \geq 0 \). ii) Deduce the least and the greates possible values of the expression \( \frac{x}{2 x^{2}-x+8} \) for all real values of \( x \)

Let's solve the problem step by step. ### Given: \[ x = \left(2x^{2} - x + 8\right)y \] This can be rewritten as: \[ y = \frac{x}{2x^{2} - x + 8} \] ### Part i) Find the values of \( y \) for which \( 1 + 2y - 63y^{2} \geq 0 \). We need to solve the quadratic inequality: \[ 1 + 2y - 63y^{2} \geq 0 \] #### Step 1: Rearrange the inequality \[ -63y^{2} + 2y + 1 \geq 0 \] #### Step 2: Multiply both sides by \(-1\) (remember to reverse the inequality sign): \[ 63y^{2} - 2y - 1 \leq 0 \] #### Step 3: Find the roots of the quadratic equation \( 63y^{2} - 2y - 1 = 0 \) using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 63 \), \( b = -2 \), and \( c = -1 \). \[ y = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 63 \times (-1)}}{2 \times 63} = \frac{2 \pm \sqrt{4 + 252}}{126} = \frac{2 \pm 16}{126} \] So, the roots are: \[ y_1 = \frac{18}{126} = \frac{3}{21} = \frac{1}{7} \] \[ y_2 = \frac{-14}{126} = -\frac{1}{9} \] #### Step 4: Determine the interval where the inequality holds Since the parabola opens upwards (coefficient of \( y^2 \) is positive), the inequality \( 63y^{2} - 2y - 1 \leq 0 \) holds between the roots. \[ -\frac{1}{9} \leq y \leq \frac{1}{7} \] ### Answer for Part i: The values of \( y \) that satisfy \( 1 + 2y - 63y^{2} \geq 0 \) are all real numbers \( y \) such that: \[ -\frac{1}{9} \leq y \leq \frac{1}{7} \] --- ### Part ii) Deduce the least and the greatest possible values of the expression \( \frac{x}{2x^{2} - x + 8} \) for all real values of \( x \). From the given relationship: \[ y = \frac{x}{2x^{2} - x + 8} \] and from Part i, we have: \[ -\frac{1}{9} \leq y \leq \frac{1}{7} \] This means that the expression \( \frac{x}{2x^{2} - x + 8} \) can **never** exceed \( \frac{1}{7} \) or be less than \( -\frac{1}{9} \) for any real value of \( x \). ### Answer for Part ii: The expression \( \frac{x}{2x^{2} - x + 8} \) attains its **greatest** value of \( \frac{1}{7} \) and its **least** value of \( -\frac{1}{9} \) for all real numbers \( x \). ### Final Summary: 1. **Values of \( y \):** \[ -\frac{1}{9} \leq y \leq \frac{1}{7} \] 2. **Extremes of \( \frac{x}{2x^{2} - x + 8} \):** - **Least value:** \( -\frac{1}{9} \) - **Greatest value:** \( \frac{1}{7} \) **Answer:** Problem i Answer: All real numbers y with –1⁄9 ≤ y ≤ 1⁄7 satisfy 1 + 2y – 63y² ≥ 0. Problem ii Answer: The expression x⁄(2x² – x + 8) has its least value –1/9 and its greatest value 1/7 for real x.