GCSE OCR Maths Solving quadratic equations: Show that the equation $x^3 + x^

Show that the equation $x^3 + x^2 - 5 = 0$ has a solution between $x = 1$ and $x = 2$ - OCR - GCSE Maths - Question 18 - 2023 - Paper 4

Question 18

Show that the equation $x^3 + x^2 - 5 = 0$ has a solution between $x = 1$ and $x = 2$. (b) Find this solution correct to 1 decimal place. You must show calculations... show full transcript

Worked Solution & Example Answer:Show that the equation $x^3 + x^2 - 5 = 0$ has a solution between $x = 1$ and $x = 2$ - OCR - GCSE Maths - Question 18 - 2023 - Paper 4

Step 1

Show that the equation $x^3 + x^2 - 5 = 0$ has a solution between $x = 1$ and $x = 2$

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Answer

To determine if there is a solution between $x = 1$ and $x = 2$ , we will evaluate the function at these points:

Let $f(x) = x^3 + x^2 - 5$ .

Calculate $f(1)$ : $f(1) = 1^3 + 1^2 - 5 = 1 + 1 - 5 = -3$
Calculate $f(2)$ : $f(2) = 2^3 + 2^2 - 5 = 8 + 4 - 5 = 7$

Since $f(1) = -3$ and $f(2) = 7$ , we observe that the function changes signs between $x=1$ and $x=2$ (from negative to positive). By the Intermediate Value Theorem, there must be at least one root in the interval $(1, 2)$ .

Step 2

Find this solution correct to 1 decimal place.

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Answer

We will use a numerical method, such as the bisection method, to find a more precise value of the root.

Evaluate the midpoint, $x = 1.5$ : $f(1.5) = (1.5)^3 + (1.5)^2 - 5 = 3.375 + 2.25 - 5 = 0.625$

Since $f(1.5) > 0$ , the root must be in the interval $(1, 1.5)$ .

Next, evaluate the midpoint again at $x = 1.25$ : $f(1.25) = (1.25)^3 + (1.25)^2 - 5 = 1.953125 + 1.5625 - 5 = -1.484375$

Since $f(1.25) < 0$ , the root is in the interval $(1.25, 1.5)$ .

Evaluate at $x = 1.4$ : $f(1.4) = (1.4)^3 + (1.4)^2 - 5 = 2.744 + 1.96 - 5 = -0.296$

Since $f(1.4) < 0$ , the root is in the interval $(1.4, 1.5)$ .

Evaluate at $x = 1.45$ : $f(1.45) = (1.45)^3 + (1.45)^2 - 5 = 3.052625 + 2.1025 - 5 = 0.155125$

Since $f(1.45) > 0$ , the root is in the interval $(1.4, 1.45)$ .

Evaluating at $x = 1.42$ : $f(1.42) = (1.42)^3 + (1.42)^2 - 5 = 2.857208 + 2.0164 - 5 = -0.126392$
Continue until we reach our required accuracy, determining the final value lies between $1.4$ and $1.5$ , correct to 1 decimal place:

Thus, the solution is approximately $1.4$ .

Show that the equation $x^3 + x^2 - 5 = 0$ has a solution between $x = 1$ and $x = 2$ - OCR - GCSE Maths - Question 18 - 2023 - Paper 4

Worked Solution & Example Answer:Show that the equation $x^3 + x^2 - 5 = 0$ has a solution between $x = 1$ and $x = 2$ - OCR - GCSE Maths - Question 18 - 2023 - Paper 4

Show that the equation $x^3 + x^2 - 5 = 0$ has a solution between $x = 1$ and $x = 2$

Find this solution correct to 1 decimal place.

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