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

Starting with the initial value $x_1 = 1$:

Calculate $x_2$ using the derived formula:

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

Now, using $x_2 = \frac{3}{4}$ to find $x_3$:

$x_3 = \frac{4(\frac{3}{4})^3 + (\frac{3}{4})^2 + 1}{6(\frac{3}{4})^2 + 2(\frac{3}{4})}$ This simplifies to:

1. Calculate the numerator:

   • $= 4 \left(\frac{27}{64}\right) + \frac{9}{16} + 1 = \frac{108}{64} + \frac{36}{64} + \frac{64}{64} = \frac{208}{64} = \frac{13}{4}$

2. Calculate the denominator:

   • $= 6 \left(\frac{9}{16}\right) + \frac{6}{4} = \frac{54}{16} + \frac{24}{16} = \frac{78}{16} = \frac{39}{8}$

Hence, $x_3 = \frac{\frac{13}{4}}{\frac{39}{8}} = \frac{13}{4} \times \frac{8}{39} = \frac{26}{39} = \frac{2}{3}$.

Thus, the values found are $x_2 = \frac{3}{4}$ and $x_3 = \frac{2}{3}$.

The Newton-Raphson method cannot be used with $x_1 = 0$ for the following reasons:

1. Substituting $x_1 = 0$ into the Newton-Raphson formula results in: $f(0) = 2(0)^2 + (0)^3 - 1 = -1$ $f'(0) = 6(0)^2 + 2(0) = 0$ This makes the denominator of the Newton-Raphson formula zero, which leads to an undefined expression.

2. Additionally, near the point $x = 0$, the function has a tangent that does not intersect the x-axis due to the characteristics of the curve of $f(x)$, indicating that convergence cannot be achieved using this method starting with $x_1 = 0$.