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

GCSE Edexcel Maths Rearranging Formulae: Show that the equation $x^2 + x

Show that the equation $x^2 + x = 7$ has a solution between 1 and 2 - Edexcel - GCSE Maths - Question 18 - 2018 - Paper 3

Question 18

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 $x = rac{ ext{√}7 - x}{2}$. St... show full transcript

Worked Solution & Example Answer:Show that the equation $x^2 + x = 7$ has a solution between 1 and 2 - Edexcel - GCSE Maths - Question 18 - 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 demonstrate that there is a solution between 1 and 2, we can evaluate the function:

$f(x) = x^2 + x - 7$

Calculating the values:

For $x = 1$ : $f(1) = 1^2 + 1 - 7 = 1 + 1 - 7 = -5$
For $x = 2$ : $f(2) = 2^2 + 2 - 7 = 4 + 2 - 7 = -1$

Since $f(1) < 0$ and $f(2) < 0$ , let's try for $x = 2.5$ :

For $x = 2.5$ : $f(2.5) = (2.5)^2 + 2.5 - 7 = 6.25 + 2.5 - 7 = 1.75$

So, now we see that $f(1) < 0$ and $f(2.5) > 0$ . Therefore, by the Intermediate Value Theorem, there is at least one root in the interval $(1, 2.5)$ hence a solution exists between 1 and 2.

Step 2

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

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Answer

Starting from the original equation,

$x^2 + x = 7$

We can rearrange it by isolating $x$ :

Subtract $x$ from both sides: $x^2 = 7 - x$
Take the square root of both sides: $x = \sqrt{7 - x}$

Thus, we have successfully rearranged the equation.

Step 3

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

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Answer

Using the iteration formula:

Let $x_0 = 2$ : $x_1 = \sqrt{7 - x_0} = \sqrt{7 - 2} = \sqrt{5} \approx 2.236$
Now, use $x_1$ to find $x_2$ : $x_2 = \sqrt{7 - x_1} = \sqrt{7 - \sqrt{5}} \approx \sqrt{7 - 2.236} \approx \sqrt{4.764} \approx 2.18$
Finally, use $x_2$ to find $x_3$ : $x_3 = \sqrt{7 - x_2} = \sqrt{7 - 2.18} \approx \sqrt{4.82} \approx 2.2$

Thus, after three iterations, we get an estimate for the solution to be approximately $x \approx 2.20$ .

Show that the equation $x^2 + x = 7$ has a solution between 1 and 2 - Edexcel - GCSE Maths - Question 18 - 2018 - Paper 3

Worked Solution & Example Answer:Show that the equation $x^2 + x = 7$ has a solution between 1 and 2 - Edexcel - GCSE Maths - Question 18 - 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 $x = \sqrt{7 - x}$

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

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