The equation $x^2 = x^3 + x - 3$ has a single solution, $x = \alpha$. 7 (a) By considering a suitable change of sign, show that $\alpha$ lies between 1.5 and 1.6 7 (b) Show that the equation $x^2 = x^3 + x - 3$ can be rearranged into the form $x^2 = x - 1 + \frac{3}{x}$ 7 (c) Use the iterative formula $x_{n+1} = \sqrt{x_n - 1 + \frac{3}{x_n}}$ with $x_1 = 1.5$, to find $x_2$, $x_3$ and $x_4$, giving your answers to four decimal places. 7 (d) Hence, deduce an interval of width 0.001 in which $\alpha$ lies.

A-Level AQA Maths Mechanics Quantities, Units & Modelling: The equation $x^2 = x^3 + x

The equation $x^2 = x^3 + x - 3$ has a single solution, $x = \alpha$ - AQA - A-Level Maths Mechanics - Question 7 - 2021 - Paper 1

Question 7

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The equation $x^2 = x^3 + x - 3$ has a single solution, $x = \alpha$. 7 (a) By considering a suitable change of sign, show that $\alpha$ lies between 1.5 and 1.6 7... show full transcript

Worked Solution & Example Answer:The equation $x^2 = x^3 + x - 3$ has a single solution, $x = \alpha$ - AQA - A-Level Maths Mechanics - Question 7 - 2021 - Paper 1

Step 1

By considering a suitable change of sign, show that $\alpha$ lies between 1.5 and 1.6

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Answer

To demonstrate that $\alpha$ lies between 1.5 and 1.6, we evaluate the function at these points:

Let ( f(x) = x^2 - (x^3 + x - 3) = -x^3 + x^2 - x + 3 )

Evaluate at $x = 1.5$ : [ f(1.5) = -(1.5)^3 + (1.5)^2 - (1.5) + 3 = -3.375 + 2.25 - 1.5 + 3 = 0.375 ]
Evaluate at $x = 1.6$ : [ f(1.6) = -(1.6)^3 + (1.6)^2 - (1.6) + 3 = -4.096 + 2.56 - 1.6 + 3 = -0.136 ]

Since $f(1.5) > 0$ and $f(1.6) < 0$ , by the Intermediate Value Theorem, there exists at least one root $\alpha$ in the interval $(1.5, 1.6)$ .

Step 2

Show that the equation $x^2 = x^3 + x - 3$ can be rearranged into the form $x^2 = x - 1 + \frac{3}{x}$

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Answer

Starting from the equation:

[ x^2 = x^3 + x - 3 ]

We can rearrange this equation:

Move all terms to one side: [ x^3 + x - x^2 - 3 = 0 ]
Isolate $x^2$ : [ x^2 = x^3 + x - 3 ]
Divide by $x$ (assuming $x \neq 0$ ): [ x^2 = x - 1 + \frac{3}{x} ]

Thus, we have demonstrated the required form.

Step 3

Use the iterative formula $x_{n+1} = \sqrt{x_n - 1 + \frac{3}{x_n}}$ with $x_1 = 1.5$, to find $x_2$, $x_3$ and $x_4$, giving your answers to four decimal places.

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Answer

Let's calculate:

For $x_1 = 1.5$ : [ x_2 = \sqrt{1.5 - 1 + \frac{3}{1.5}} = \sqrt{0.5 + 2} = \sqrt{2.5} \approx 1.5811 ]
For $x_2 \approx 1.5811$ : [ x_3 = \sqrt{1.5811 - 1 + \frac{3}{1.5811}} \approx \sqrt{0.5811 + 1.8975} = \sqrt{2.4786} \approx 1.5743 ]
For $x_3 \approx 1.5743$ : [ x_4 = \sqrt{1.5743 - 1 + \frac{3}{1.5743}} \approx \sqrt{0.5743 + 1.9053} = \sqrt{2.4796} \approx 1.5748 ]

Thus: $x_2 \approx 1.5811$ , $x_3 \approx 1.5743$ , $x_4 \approx 1.5748$ .

Step 4

Hence, deduce an interval of width 0.001 in which $\alpha$ lies.

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Answer

From our iterative results, we found:

$x_3 \approx 1.5743$
$x_4 \approx 1.5748$

The values are consistent, leading us to deduce that:

The value of $\alpha$ lies within the interval $[1.5743, 1.5748]$ . To create an interval of width 0.001, we can state:

[ 1.5743 \leq \alpha \leq 1.5748 ]

Therefore, an interval of width 0.001 in which $\alpha$ lies is $[1.5743, 1.5753]$ .

The equation $x^2 = x^3 + x - 3$ has a single solution, $x = \alpha$ - AQA - A-Level Maths Mechanics - Question 7 - 2021 - Paper 1

Worked Solution & Example Answer:The equation $x^2 = x^3 + x - 3$ has a single solution, $x = \alpha$ - AQA - A-Level Maths Mechanics - Question 7 - 2021 - Paper 1

By considering a suitable change of sign, show that $\alpha$ lies between 1.5 and 1.6

Show that the equation $x^2 = x^3 + x - 3$ can be rearranged into the form $x^2 = x - 1 + \frac{3}{x}$

Use the iterative formula $x_{n+1} = \sqrt{x_n - 1 + \frac{3}{x_n}}$ with $x_1 = 1.5$, to find $x_2$, $x_3$ and $x_4$, giving your answers to four decimal places.

Hence, deduce an interval of width 0.001 in which $\alpha$ lies.

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