The equation $2x^2 + x^3 - 1 = 0$ has exactly one real root. (a) Show that, for this equation, the Newton-Raphson formula can be written $$x_{n+1} = \frac{4x_n^3 + x_n^2 + 1}{6x_n^2 + 2x_n}$$ (3) Using the formula given in part (a) with $x_1 = 1$ (b) find the values of $x_2$ and $x_3$ (2) (c) Explain why, for this question, the Newton-Raphson method cannot be used with $x_1 = 0$ (1)

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The equation $2x^2 + x^3 - 1 = 0$ has exactly one real root - Edexcel - A-Level Maths Pure - Question 7 - 2018 - Paper 2

Question 7

The equation $2x^2 + x^3 - 1 = 0$ has exactly one real root. (a) Show that, for this equation, the Newton-Raphson formula can be written $$x_{n+1} = \frac{4x_n^3 + ... show full transcript

Worked Solution & Example Answer:The equation $2x^2 + x^3 - 1 = 0$ has exactly one real root - Edexcel - A-Level Maths Pure - Question 7 - 2018 - Paper 2

Step 1

Show that, for this equation, the Newton-Raphson formula can be written

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Answer

To derive the Newton-Raphson formula, we start by noting the function: $f(x) = 2x^2 + x^3 - 1$

The derivative is: $f'(x) = 6x^2 + 2x$

In the Newton-Raphson formula, $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

Substituting gives: $x_{n+1} = x_n - \frac{2x_n^2 + x_n^3 - 1}{6x_n^2 + 2x_n}$

To fix this into the required form, we simplify: $x_{n+1} = \frac{4x_n^3 + x_n^2 + 1}{6x_n^2 + 2x_n}$ .

Step 2

find the values of $x_2$ and $x_3$

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Answer

Using the formula with $x_1 = 1$ :

First, calculate $x_2$ : $x_2 = \frac{4(1)^3 + (1)^2 + 1}{6(1)^2 + 2(1)} = \frac{4 + 1 + 1}{6 + 2} = \frac{6}{8} = \frac{3}{4}$

Next, calculate $x_3$ using $x_2$ : $x_3 = \frac{4(\frac{3}{4})^3 + (\frac{3}{4})^2 + 1}{6(\frac{3}{4})^2 + 2(\frac{3}{4})}$

Calculating further, we get: $x_3 = \frac{4(\frac{27}{64}) + (\frac{9}{16}) + 1}{6(\frac{9}{16}) + 2(\frac{3}{4})} = \frac{\frac{108}{64} + \frac{36}{64} + \frac{64}{64}}{\frac{54}{16} + \frac{24}{16}} = \frac{\frac{208}{64}}{\frac{78}{16}}$

Simplifying results in: $x_3 = \frac{208}{64} \times \frac{16}{78} = \frac{52}{39} = \frac{4}{3}$ .

Step 3

Explain why, for this question, the Newton-Raphson method cannot be used with $x_1 = 0$

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Answer

The Newton-Raphson method cannot be used with $x_1 = 0$ because:

At $x = 0$ , the function value $f(0) = 2(0)^2 + (0)^3 - 1 = -1$ and its derivative $f'(0) = 6(0)^2 + 2(0) = 0$ . Thus, the formula would attempt to divide by zero.
Therefore, this point is not suitable as it leads to an undefined operation.

The equation $2x^2 + x^3 - 1 = 0$ has exactly one real root - Edexcel - A-Level Maths Pure - Question 7 - 2018 - Paper 2

Worked Solution & Example Answer:The equation $2x^2 + x^3 - 1 = 0$ has exactly one real root - Edexcel - A-Level Maths Pure - Question 7 - 2018 - Paper 2

Show that, for this equation, the Newton-Raphson formula can be written

find the values of $x_2$ and $x_3$

Explain why, for this question, the Newton-Raphson method cannot be used with $x_1 = 0$

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