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Using $x_{nt} = -2 - \frac{4}{x_{t}}$ with $x_{t} = -2.5$ - Edexcel - GCSE Maths - Question 16 - 2017 - Paper 3

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Using-$x_{nt}-=--2---\frac{4}{x_{t}}$-with-$x_{t}-=--2.5$-Edexcel-GCSE Maths-Question 16-2017-Paper 3.png

Using $x_{nt} = -2 - \frac{4}{x_{t}}$ with $x_{t} = -2.5$. (a) find the values of $x_{t}$, $x_{1}$, and $x_{s}$. $x_{1} = $ $x_{s} = $ (b) Explain the relationsh... show full transcript

Worked Solution & Example Answer:Using $x_{nt} = -2 - \frac{4}{x_{t}}$ with $x_{t} = -2.5$ - Edexcel - GCSE Maths - Question 16 - 2017 - Paper 3

Step 1

find the values of $x_{1}$, $x_{s}$, and $x_{t}$

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Answer

To find the values, we start with the given equation:

xnt=−2−4xtx_{nt} = -2 - \frac{4}{x_{t}}.

Substituting xt=−2.5x_{t} = -2.5 into the equation:

xnt=−2−4−2.5x_{nt} = -2 - \frac{4}{-2.5}

Calculating this gives:

xnt=−2+1.6=−0.4x_{nt} = -2 + 1.6 = -0.4

Thus, we have:

  • x1=−0.4x_{1} = -0.4
  • The values for xsx_{s} and others can be derived from further substitutions or iterations, depending on the method of solving.

Step 2

Explain the relationship between the values of $x_{1}$, $x_{t}$, and $x_{s}$

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Answer

The values x1x_{1}, xtx_{t}, and xsx_{s} represent estimations of the solution for the equation x2+2x+4=0x^{2} + 2x + 4 = 0. While xtx_{t} can be viewed as an initial approximation, x1x_{1} refines this estimate further.

The equation has no real solutions as its discriminant, calculated as b2−4ac=22−4(1)(4)=4−16=−12b^2 - 4ac = 2^2 - 4(1)(4) = 4 - 16 = -12, is negative. The values are essentially representing iterative methods approaching a complex solution, highlighting the relationship between algebraic formulations and numerical approximations.

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