16 (a) Use the iteration formula $x_{n} = \\sqrt{10 - 2x_{n-1}}$, to find the values of $x_{1}, x_{2}$, and $x_{3}$ - Edexcel - GCSE Maths - Question 18 - 2021 - Paper 2

Now, using $x_{1}$ to find $x_{2}$:

1. Compute: $x_{2} = \\sqrt{10 - 2 \cdot 2.449} = \\sqrt{5.102} \approx 2.26.$

Now, using $x_{2}$ to find $x_{3}$:

1. Compute: $x_{3} = \\sqrt{10 - 2 \cdot 2.26} = \\sqrt{5.48} \approx 2.34.$

Given the values of $x_{1} \approx 2.45$, $x_{2} \approx 2.26$, and $x_{3} \approx 2.34$, we can see that these values are close to the root of the quadratic equation:

• Assume $x^{2} + ax + b = 0$ with roots being $x_{1}, x_{2}$, and $x_{3}$.
• We know:
  • The sum of roots $x_{1} + x_{2} + x_{3} = -a$.
  • The product of roots $x_{1} \cdot x_{2} + x_{1} \cdot x_{3} + x_{2} \cdot x_{3} = b$.

Calculating:

1. The sum is approximately: $x_{1} + x_{2} + x_{3} \approx 2.45 + 2.26 + 2.34 \approx 7.05,$ therefore, $-a \approx 7.05 \Rightarrow a \approx -7.$

2. The product gives: $x_{1} \cdot x_{2} + x_{1} \cdot x_{3} + x_{2} \cdot x_{3} \approx 2.45 \cdot 2.26 + 2.45 \cdot 2.34 + 2.26 \cdot 2.34 \approx 5.54 + 5.74 + 5.28 \approx 16.56,$ thus $b \approx 16.$