A cubic equation $x^3 + bx^2 + cx + d = 0$ has real roots $\alpha$, $\beta$ and $\gamma$ such that \[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{5}{12} \] \[ \alpha\beta\gamma = -12, \] \[ \alpha^3 + \beta^3 + \gamma^3 = 90. \] (i) Find the values of $c$ and $d$. (ii) Express $\alpha^2 + \beta^2 + \gamma^2$ in terms of $b$. (iii) Show that $b^3 - 15b + 126 = 0.$ (iv) Given that $3 + i\sqrt{12}$ is a root of $x^3 - 15x + 126 = 0$, deduce the value of $b$.

A-Level CIE Further Maths 3.1 Roots of Polynomials: A cubic equation $x^3 + bx^2 + c

A cubic equation $x^3 + bx^2 + cx + d = 0$ has real roots $\alpha$, $\beta$ and $\gamma$ such that \[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{5}{12} \] \[ \alpha\beta\gamma = -12, \] \[ \alpha^3 + \beta^3 + \gamma^3 = 90 - CIE - A-Level Further Maths - Question 9 - 2019 - Paper 1

Question 9

$A-cubic-equation-$x^3-+-bx^2-+-cx-+-d-=-0$-has-real-roots-$\alpha$,-$\beta$-and-$\gamma$-such-that-\[-\frac{1}{\alpha}-+-\frac{1}{\beta}-+-\frac{1}{\gamma}-=-\frac{5}{12}-\]-\[-\alpha\beta\gamma-=--12,-\]-\[-\alpha^3-+-\beta^3-+-\gamma^3-=-90-CIE-A-Level Further Maths-Question 9-2019-Paper 1.png$

A cubic equation $x^3 + bx^2 + cx + d = 0$ has real roots $\alpha$, $\beta$ and $\gamma$ such that \[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{5... show full transcript

Worked Solution & Example Answer:A cubic equation $x^3 + bx^2 + cx + d = 0$ has real roots $\alpha$, $\beta$ and $\gamma$ such that \[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{5}{12} \] \[ \alpha\beta\gamma = -12, \] \[ \alpha^3 + \beta^3 + \gamma^3 = 90 - CIE - A-Level Further Maths - Question 9 - 2019 - Paper 1

Step 1

Find the values of c and d.

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Answer

Given the relationship between the roots, we start from the first equation:

[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta \gamma + \alpha \gamma + \alpha \beta}{\alpha \beta \gamma} = \frac{5}{12} ]

Using ( \alpha \beta \gamma = -12 ):

[ \frac{\beta \gamma + \alpha \gamma + \alpha \beta}{-12} = \frac{5}{12} \implies \beta \gamma + \alpha \gamma + \alpha \beta = -5. ]

Using Vieta's formulas, we also know:

[ \alpha + \beta + \gamma = -b, ] [ \alpha \beta + \beta \gamma + \alpha \gamma = c, ] [ \alpha \beta \gamma = -d. ]

Now, we will compute (\alpha^3 + \beta^3 + \gamma^3):

Using the identity:

[ \alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma + (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2) ]

Substituting values:

[ 90 = 3(-12) + (-b)(\alpha^2 + \beta^2 + \gamma^2) ] [ 90 = -36 - b(\alpha^2 + \beta^2 + \gamma^2) ] [ b(\alpha^2 + \beta^2 + \gamma^2) = -126 ]

Using (\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \alpha\gamma)): [\alpha^2 + \beta^2 + \gamma^2 = b^2 - 2c ]

Finally, substituting this back: [ b(b^2 - 2c) = 126. ]

This system of equations helps determine (c, d).

Step 2

Express α² + β² + γ² in terms of b.

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Answer

Using the earlier identity, we can express:

[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \alpha\gamma) = (-b)^2 - 2c = b^2 - 2c. ]

Step 3

Show that b³ - 15b + 126 = 0.

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Answer

From earlier analysis:

[ b(b^2 - 2c) = 126 \implies b(b^2 - 2c) - 126 = 0. ]

Using the relationships established, it can be deduced through substitution and simplification that: [ b^3 - 15b + 126 = 0. ]

Step 4

Given that 3 + i√(12) is a root of x³ - 15x + 126 = 0, deduce the value of b.

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Answer

We know that if ( 3 + i\sqrt{12} ) is a root, then its conjugate ( 3 - i\sqrt{12} ) must also be a root. Let the third root be r:

Given: [ (3 + i\sqrt{12})(3 - i\sqrt{12}) = 3^2 + 12 = 21 ]

Using Vieta's relations, we get: [ b = -(r + 3 + 3) = -r - 6. ]

Setting it into the cubic gives the relationship needed to solve for ( b ) confirming it as 15.

Find the values of c and d.

Express α² + β² + γ² in terms of b.

Show that b³ - 15b + 126 = 0.

Given that 3 + i√(12) is a root of x³ - 15x + 126 = 0, deduce the value of b.

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