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Show that the equation $x^4 - x^2 - 9 = 0$ has a solution between $x = 1$ and $x = 2$ - OCR - GCSE Maths - Question 20 - 2019 - Paper 6

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Show-that-the-equation-$x^4---x^2---9-=-0$-has-a-solution-between-$x-=-1$-and-$x-=-2$-OCR-GCSE Maths-Question 20-2019-Paper 6.png

Show that the equation $x^4 - x^2 - 9 = 0$ has a solution between $x = 1$ and $x = 2$. Find this solution correct to 1 decimal place. Show your working.

Worked Solution & Example Answer:Show that the equation $x^4 - x^2 - 9 = 0$ has a solution between $x = 1$ and $x = 2$ - OCR - GCSE Maths - Question 20 - 2019 - Paper 6

Step 1

Show that the equation $x^4 - x^2 - 9 = 0$ has a solution between $x = 1$ and $x = 2$

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Answer

To verify if there is a solution between x=1x = 1 and x=2x = 2, we will evaluate the function at these two points.

Let: f(x)=x4x29f(x) = x^4 - x^2 - 9

Calculate f(1)f(1): f(1)=14129=119=9f(1) = 1^4 - 1^2 - 9 = 1 - 1 - 9 = -9

Calculate f(2)f(2): f(2)=24229=1649=3f(2) = 2^4 - 2^2 - 9 = 16 - 4 - 9 = 3

Since f(1)=9f(1) = -9 (a negative value) and f(2)=3f(2) = 3 (a positive value), by the Intermediate Value Theorem, there exists at least one root in the interval (1,2)(1, 2).

Step 2

Find this solution correct to 1 decimal place. Show your working.

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Answer

To find the solution, we will use the method of bisection.

Let’s evaluate the midpoint: x=1.5x = 1.5 Calculating f(1.5)f(1.5): f(1.5)=(1.5)4(1.5)29=5.06252.259=6.1875f(1.5) = (1.5)^4 - (1.5)^2 - 9 = 5.0625 - 2.25 - 9 = -6.1875

As f(1.5)<0f(1.5) < 0, the root lies in (1.5,2)(1.5, 2).

Next, evaluate the midpoint of (1.5,2)(1.5, 2), which is: x=1.75x = 1.75 Calculating f(1.75)f(1.75): f(1.75)=(1.75)4(1.75)29=9.37893.06259=2.6836f(1.75) = (1.75)^4 - (1.75)^2 - 9 = 9.3789 - 3.0625 - 9 = -2.6836

Since f(1.75)<0f(1.75) < 0, the root lies in (1.75,2)(1.75, 2).

Now evaluate: x=1.875x = 1.875 Calculating f(1.875)f(1.875): f(1.875)=(1.875)4(1.875)29=11.41023.51569=1.1054f(1.875) = (1.875)^4 - (1.875)^2 - 9 = 11.4102 - 3.5156 - 9 = -1.1054

Continuing in this manner, we evaluate: x=1.9375x = 1.9375 Calculating f(1.9375)f(1.9375): f(1.9375)=(1.9375)4(1.9375)29=12.86643.74619=0.1203f(1.9375) = (1.9375)^4 - (1.9375)^2 - 9 = 12.8664 - 3.7461 - 9 = 0.1203

Now, since f(1.875)<0f(1.875) < 0 and f(1.9375)>0f(1.9375) > 0, the solution lies in (1.875,1.9375)(1.875, 1.9375).

Repeating this process leads us to narrow down to: x=1.9x = 1.9 Calculating: f(1.9)=1.941.929=12.03493.619=0.5751f(1.9) = 1.9^4 - 1.9^2 - 9 = 12.0349 - 3.61 - 9 = -0.5751

And finally evaluating: f(1.91)=12.11823.66819=0.5499f(1.91) = 12.1182 - 3.6681 - 9 = -0.5499

Concluding that the solution is approximately: x=1.9x = 1.9

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