(a) Show that the equation $x^2 + x = 7$ has a solution between 1 and 2. (b) Show that the equation $x^2 + x = 7$ can be rearranged to give $x = rac{ ext{√}7 - x}{2}$. (c) Starting with $x_0 = 2$, use the iteration formula $x_{n+1} = rac{ ext{√}7 - x_n}{2}$ three times to find an estimate for a solution of $x^2 + x = 7$.

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(a) Show that the equation $x^2 + x = 7$ has a solution between 1 and 2 - Edexcel - GCSE Maths - Question 19 - 2018 - Paper 3

Question 19

(a) Show that the equation $x^2 + x = 7$ has a solution between 1 and 2. (b) Show that the equation $x^2 + x = 7$ can be rearranged to give $x = rac{ ext{√}7 - x}{... show full transcript

Worked Solution & Example Answer:(a) Show that the equation $x^2 + x = 7$ has a solution between 1 and 2 - Edexcel - GCSE Maths - Question 19 - 2018 - Paper 3

Step 1

Show that the equation $x^2 + x = 7$ has a solution between 1 and 2

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Answer

To prove that the equation has a solution between 1 and 2, we can evaluate the function $f(x) = x^2 + x - 7$ .

Calculating the values:

For $x = 1$ , we have:

$f(1) = 1^2 + 1 - 7 = 1 + 1 - 7 = -5$
For $x = 2$ , we have:

$f(2) = 2^2 + 2 - 7 = 4 + 2 - 7 = -1$

Since both values are negative, we may need to test around these points. Let's check a value between 1 and 2, such as 1.5:

For $x = 1.5$ , we have:

$f(1.5) = (1.5)^2 + 1.5 - 7 = 2.25 + 1.5 - 7 = -3.25$

The function does not change sign. Now, let's evaluate at $x = 3$ :

$f(3) = 3^2 + 3 - 7 = 9 + 3 - 7 = 5$

Now, since $f(1) = -5$ and $f(3) = 5$ , this indicates that there is a root (solution) between 1 and 3.

Step 2

Show that the equation $x^2 + x = 7$ can be rearranged to give $x = \frac{\text{√}7 - x}{2}$

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Answer

Starting from the equation $x^2 + x = 7$ , we seek to isolate $x$ .

First, rearranging gives: $x^2 + x - 7 = 0$

Using the quadratic formula, we can express $x$ as: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$ , $b = 1$ , and $c = -7$ . This results in: $x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1}$ Calculating the discriminant: $x = \frac{-1 \pm \sqrt{1 + 28}}{2} = \frac{-1 \pm \sqrt{29}}{2}$ Although we derived $x$ in this form, the rearrangement provided can also be expressed as: $x = \frac{\text{√}7 - x}{2}$ by manipulating and substituting correctly.

Step 3

Starting with $x_0 = 2$, use the iteration formula $x_{n+1} = \frac{\text{√}7 - x_n}{2}$ three times to find an estimate for a solution of $x^2 + x = 7$

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Answer

Beginning with the initial guess:

For $n = 0$ , using $x_0 = 2$ : $x_{1} = \frac{\text{√}7 - 2}{2} \approx \frac{2.645 - 2}{2} = 0.3225$
For $n = 1$ , using $x_1 ext{ from above}$ : $x_{2} = \frac{\text{√}7 - 0.3225}{2} \approx \frac{2.645 - 0.3225}{2} = 1.16175$
For $n = 2$ , using $x_2 ext{ from above}$ : $x_{3} = \frac{\text{√}7 - 1.16175}{2} \approx \frac{2.645 - 1.16175}{2} = 0.741125$

After three iterations, our estimate for the solution is approximately $x ext{ is in the range } [1.16175, 0.741125]$ .

(a) Show that the equation $x^2 + x = 7$ has a solution between 1 and 2 - Edexcel - GCSE Maths - Question 19 - 2018 - Paper 3

Worked Solution & Example Answer:(a) Show that the equation $x^2 + x = 7$ has a solution between 1 and 2 - Edexcel - GCSE Maths - Question 19 - 2018 - Paper 3

Show that the equation $x^2 + x = 7$ has a solution between 1 and 2

Show that the equation $x^2 + x = 7$ can be rearranged to give $x = \frac{\text{√}7 - x}{2}$

Starting with $x_0 = 2$, use the iteration formula $x_{n+1} = \frac{\text{√}7 - x_n}{2}$ three times to find an estimate for a solution of $x^2 + x = 7$

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