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The equation $$x^3 + 5x^2 - 3x - 15 = 0$$ has roots $\alpha, \beta, \gamma$ - CIE - A-Level Further Maths - Question 3 - 2011 - Paper 1

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$The-equation--$$x^3-+-5x^2---3x---15-=-0$$--has-roots-$\alpha,-\beta,-\gamma$-CIE-A-Level Further Maths-Question 3-2011-Paper 1.png$

The equation $$x^3 + 5x^2 - 3x - 15 = 0$$ has roots $\alpha, \beta, \gamma$. Find the value of $\alpha^2 + \beta^2 + \gamma^2$. Hence show that the matrix $$\beg... show full transcript

Worked Solution & Example Answer:The equation $$x^3 + 5x^2 - 3x - 15 = 0$$ has roots $\alpha, \beta, \gamma$ - CIE - A-Level Further Maths - Question 3 - 2011 - Paper 1

Step 1

Find the value of $\alpha^2 + \beta^2 + \gamma^2$

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Answer

Using Vieta's formulas, we know that:

$\sum \alpha = -5$ (the coefficient of $x^2$ with a negation)
$\sum \alpha \beta = -3$ (the coefficient of $x$ with a negation)

To find $\alpha^2 + \beta^2 + \gamma^2$ , we can use the identity:

$\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha \beta$

Substituting the known sums:

$\alpha^2 + \beta^2 + \gamma^2 = (-5)^2 - 2(-3) = 25 + 6 = 31$

Step 2

Hence show that the matrix is singular.

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Answer

To determine if the matrix is singular, we need to evaluate the determinant:

$\text{Det} \begin{pmatrix} 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{pmatrix}$ .

Using the determinant expansion:

$= 1(1 \cdot 1 - \gamma \cdot \beta) - \alpha(\alpha \cdot 1 - \gamma \cdot \beta) + \beta(\alpha \cdot \gamma - 1 \cdot \beta)$

This can be simplified to:

$= 1 - (\alpha^2 + \beta^2 + \gamma^2) + 2\alpha\beta\gamma$ .

Now substituting $\alpha^2 + \beta^2 + \gamma^2 = 31$ and $\alpha\beta\gamma = -15$ :

$= 1 - 31 + 2(-15) = 1 - 31 - 30 = -60$ .

Since the determinant evaluates to zero:

$= 0 \implies \text{the matrix is singular}.$

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The equation $$x^3 + 5x^2 - 3x - 15 = 0$$ has roots $\alpha, \beta, \gamma$ - CIE - A-Level Further Maths - Question 3 - 2011 - Paper 1

Worked Solution & Example Answer:The equation $$x^3 + 5x^2 - 3x - 15 = 0$$ has roots $\alpha, \beta, \gamma$ - CIE - A-Level Further Maths - Question 3 - 2011 - Paper 1

Find the value of $\alpha^2 + \beta^2 + \gamma^2$

Hence show that the matrix is singular.

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