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The equation $x^3 - 3x + 1 = 0$ has three real roots - AQA - A-Level Maths Pure - Question 4 - 2017 - Paper 2

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The equation $x^3 - 3x + 1 = 0$ has three real roots. (a) Show that one of the roots lies between -2 and -1. (b) Taking $x_1 = -2$ as the first approximation to on... show full transcript

Worked Solution & Example Answer:The equation $x^3 - 3x + 1 = 0$ has three real roots - AQA - A-Level Maths Pure - Question 4 - 2017 - Paper 2

Step 1

Show that one of the roots lies between -2 and -1

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Answer

To show that one of the roots lies between -2 and -1, we will evaluate the function:

f(x)=x3−3x+1f(x) = x^3 - 3x + 1

Evaluate at x=−2x = -2:

f(−2)=(−2)3−3(−2)+1=−8+6+1=−1f(-2) = (-2)^3 - 3(-2) + 1 = -8 + 6 + 1 = -1

Now, evaluate at x=−1x = -1:

f(−1)=(−1)3−3(−1)+1=−1+3+1=3f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3

Since f(−2)=−1<0f(-2) = -1 < 0 and f(−1)=3>0f(-1) = 3 > 0, and by the Intermediate Value Theorem, there is at least one root between -2 and -1.

Step 2

Taking x_1 = -2 as the first approximation to one of the roots, use the Newton-Raphson method to find x_2, the second approximation.

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The Newton-Raphson formula is given by:

xn+1=xn−f(xn)f′(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

Where the derivative is:

f′(x)=3x2−3f'(x) = 3x^2 - 3

Substituting x1=−2x_1 = -2 into the formula:

Evaluate f(−2)f(-2) (already computed): f(−2)=−1f(-2) = -1

Now compute f′(−2)f'(-2):

f′(−2)=3(−2)2−3=12−3=9f'(-2) = 3(-2)^2 - 3 = 12 - 3 = 9

Now apply the Newton-Raphson formula:

x2=−2−−19=−2+19=−189+19=−179x_2 = -2 - \frac{-1}{9} = -2 + \frac{1}{9} = -\frac{18}{9} + \frac{1}{9} = -\frac{17}{9}

Hence, x2=−179x_2 = \frac{-17}{9} or approximately -1.89.

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