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Using $x_{t} = -2 - \frac{4}{x_{t}}$ with $x_{t} = -2.5$ (a) find the values of $x_{t}$, $x_{1}$, and $x_{2}$ - Edexcel - GCSE Maths - Question 16 - 2017 - Paper 3

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Using-$x_{t}-=--2---\frac{4}{x_{t}}$-with-$x_{t}-=--2.5$--(a)-find-the-values-of-$x_{t}$,-$x_{1}$,-and-$x_{2}$-Edexcel-GCSE Maths-Question 16-2017-Paper 3.png

Using $x_{t} = -2 - \frac{4}{x_{t}}$ with $x_{t} = -2.5$ (a) find the values of $x_{t}$, $x_{1}$, and $x_{2}$. $x_{1} = $ $x_{2} = $ $x_{3} = $ (b) Explain the ... show full transcript

Worked Solution & Example Answer:Using $x_{t} = -2 - \frac{4}{x_{t}}$ with $x_{t} = -2.5$ (a) find the values of $x_{t}$, $x_{1}$, and $x_{2}$ - Edexcel - GCSE Maths - Question 16 - 2017 - Paper 3

Step 1

Explain the relationship between the values of $x_{1}$, $x_{2}$, and $x_{t}$, and the equation $x^{2} + 2x + 4 = 0$

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Answer

The values of x1x_{1}, x2x_{2}, and xtx_{t} are related through the iterative method of solving the equation. In particular, they represent successive approximations of a solution to the quadratic equation x2+2x+4=0x^{2} + 2x + 4 = 0. The equation has complex solutions given that its discriminant (b24ac=224(1)(4)<0b^2 - 4ac = 2^2 - 4(1)(4) < 0) is negative.

Thus, x1x_{1} and subsequent iterations are estimates that converge to one of these complex solutions. The iterative process helps in estimating solutions for nonlinear equations and illustrates how different xx values lead back to the original equation.

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