Photo AI

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 16 - 2021 - Paper 2

Question icon

Question 16

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 16-2021-Paper 2.png

16 (a) Use the iteration formula $x_{n} = \sqrt{10 - 2x_{n-1}}$ to find the values of $x_{1}$, $x_{2}$, and $x_{3}$. Start with $x_{0} = 2$. $x_{1} = $ $x_{2} = $ ... show full transcript

Worked Solution & Example Answer: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 16 - 2021 - Paper 2

Step 1

Use the iteration formula $x_{n} = \sqrt{10 - 2x_{n-1}}$ to find the values of $x_{1}$, $x_{2}$, and $x_{3}$

96%

114 rated

Answer

Starting with x0=2x_{0} = 2:

  1. Calculate x1x_{1}: x1=102×2=62.449x_{1} = \sqrt{10 - 2 \times 2} = \sqrt{6} \approx 2.449

  2. Calculate x2x_{2}: x2=102×2.449=104.898=5.1022.26x_{2} = \sqrt{10 - 2 \times 2.449} = \sqrt{10 - 4.898} = \sqrt{5.102} \approx 2.26

  3. Calculate x3x_{3}: x3=102×2.26=104.52=5.482.34x_{3} = \sqrt{10 - 2 \times 2.26} = \sqrt{10 - 4.52} = \sqrt{5.48} \approx 2.34

Thus,

  • x12.45x_{1} \approx 2.45
  • x22.26x_{2} \approx 2.26
  • x32.34x_{3} \approx 2.34

Step 2

Find the value of $a$ and the value of $b$

99%

104 rated

Answer

Using the obtained values, we know:

  • x12.45x_{1} \approx 2.45, x22.26x_{2} \approx 2.26, x32.34x_{3} \approx 2.34. We can form the quadratic equation from these values. From the quadratic form x2+ax+b=0x^{2} + ax + b = 0:

By substituting x1x_1 and x2x_2 into the equation: a=(x1+x2)(2.45+2.26)=4.715a = - (x_{1} + x_{2}) \approx - (2.45 + 2.26) = - 4.71 \approx -5 b=x1×x22.45×2.265.546b = x_{1} \times x_{2} \approx 2.45 \times 2.26 \approx 5.54 \approx 6

Hence,

  • a5a \approx -5
  • b6b \approx 6

Join the GCSE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;