The roots of the cubic equation $2x^3 + x^2 - 7 = 0$ are $\alpha$, $\beta$ and $\gamma$. Using the substitution $y = 1 + \frac{1}{x}$, or otherwise, find the cubic equation whose roots are $1 + \frac{1}{\alpha}$, $1 + \frac{1}{\beta}$ and $1 + \frac{1}{\gamma}$, giving your answer in the form $ay^3 + by^2 + cy + d = 0$, where $a$, $b$, $c$ and $d$ are constants to be found.

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The roots of the cubic equation $2x^3 + x^2 - 7 = 0$ are $\alpha$, $\beta$ and $\gamma$ - CIE - A-Level Further Maths - Question 1 - 2016 - Paper 1

Question 1

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The roots of the cubic equation $2x^3 + x^2 - 7 = 0$ are $\alpha$, $\beta$ and $\gamma$. Using the substitution $y = 1 + \frac{1}{x}$, or otherwise, find the cubic e... show full transcript

Worked Solution & Example Answer:The roots of the cubic equation $2x^3 + x^2 - 7 = 0$ are $\alpha$, $\beta$ and $\gamma$ - CIE - A-Level Further Maths - Question 1 - 2016 - Paper 1

Step 1

Using the substitution $y = 1 + \frac{1}{x}$

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Answer

We start with the substitution:

$y = 1 + \frac{1}{x} \Rightarrow x = \frac{1}{y - 1}$

Substituting into the original equation:

$2\left(\frac{1}{y - 1}\right)^3 + \left(\frac{1}{y - 1}\right)^2 - 7 = 0$

Simplifying this gives us:

$2(y - 1)^3 + (y - 1)^2 - 7(y - 1) = 0$

Expanding and simplifying terms will form a new cubic equation in $y$ .

Step 2

Finding the new cubic equation

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Answer

After substitution, we have:

$2(y - 1)^3 + (y - 1)^2 - 7(y - 1) = 0$

This expands to:

$2(y^3 - 3y^2 + 3y - 1) + (y^2 - 2y + 1) - 7y + 7 = 0$

Combining like terms yields:

$2y^3 - 6y^2 + 6y - 2 + y^2 - 2y + 1 - 7y + 7 = 0$

Thus,

$2y^3 - 5y^2 - 3y + 6 = 0$

This can be simplified further to determine $a$ , $b$ , $c$ , and $d$ as follows: $a = 2, b = -5, c = -3, d = 6.$

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The roots of the cubic equation $2x^3 + x^2 - 7 = 0$ are $\alpha$, $\beta$ and $\gamma$ - CIE - A-Level Further Maths - Question 1 - 2016 - Paper 1

Worked Solution & Example Answer:The roots of the cubic equation $2x^3 + x^2 - 7 = 0$ are $\alpha$, $\beta$ and $\gamma$ - CIE - A-Level Further Maths - Question 1 - 2016 - Paper 1

Using the substitution $y = 1 + \frac{1}{x}$

Finding the new cubic equation

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