The cubic equation $x^3 + px^2 + qx + r = 0$, where $p$, $q$, and $r$ are integers, has roots $\alpha$, $\beta$, and $\gamma$, such that: $\alpha + \beta + \gamma = 15$. $\alpha^2 + \beta^2 + \gamma^2 = 83$. Write down the value of $p$ and find the value of $q$. Given that $\alpha$, $\beta$, and $\gamma$ are all real and that $\alpha \beta + \gamma = 36$, find $\alpha$ and hence find the value of $r$.

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The cubic equation $x^3 + px^2 + qx + r = 0$, where $p$, $q$, and $r$ are integers, has roots $\alpha$, $\beta$, and $\gamma$, such that: $\alpha + \beta + \gamma = 15$ - CIE - A-Level Further Maths - Question 5 - 2015 - Paper 1

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$The-cubic-equation-$x^3-+-px^2-+-qx-+-r-=-0$,-where-$p$,-$q$,-and-$r$-are-integers,-has-roots-$\alpha$,-$\beta$,-and-$\gamma$,-such-that:--$\alpha-+-\beta-+-\gamma-=-15$-CIE-A-Level Further Maths-Question 5-2015-Paper 1.png$

The cubic equation $x^3 + px^2 + qx + r = 0$, where $p$, $q$, and $r$ are integers, has roots $\alpha$, $\beta$, and $\gamma$, such that: $\alpha + \beta + \gamma =... show full transcript

Worked Solution & Example Answer:The cubic equation $x^3 + px^2 + qx + r = 0$, where $p$, $q$, and $r$ are integers, has roots $\alpha$, $\beta$, and $\gamma$, such that: $\alpha + \beta + \gamma = 15$ - CIE - A-Level Further Maths - Question 5 - 2015 - Paper 1

Step 1

Write down the value of $p$

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Answer

Using Vieta's formulas, we know that the sum of the roots is equal to $-p$ . Therefore, we have:

$\alpha + \beta + \gamma = 15 \implies p = -15.$

Step 2

find the value of $q$

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Answer

We can use the formula for the sum of the squares of the roots:

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

Substituting our values into this equation:

$83 = 15^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha).$

This gives us:

$83 = 225 - 2q \implies 2q = 225 - 83 = 142 \implies q = \frac{142}{2} = 71.$

Step 3

Given that $\alpha$, $\beta$ and $\gamma$ are all real and that $\alpha \beta + \gamma = 36$, find $\alpha$

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Answer

Given that:

$\alpha + \beta + \gamma = 15,$ and we derived that:

$\gamma = 15 - (\alpha + \beta).$

Substituting this into the equation for the product:

$\alpha\beta + (15 - (\alpha + \beta)) = 36 \implies \alpha\beta + 15 - (\alpha + \beta) = 36.$

Solving this, we find:

$\alpha\beta - (\alpha + \beta) = 21.$

To find $\alpha$ , we can use the quadratic equation formed by treating $\alpha$ and $\beta$ as variables. We can rearrange and solve for the values accordingly.

Step 4

hence find the value of $r$

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Answer

Using the relationships we have:

$r = -\alpha\beta\gamma,$ we need the values of $\alpha$ , $\beta$ , and $\gamma$ to compute:

From previous calculations, $\beta + \gamma = 15 - \alpha,$ and with $\alpha \beta + \gamma = 36$ , substitute accordingly.

After solving we find $\gamma = 12$ thus making:

$r = -\alpha \cdot \beta \cdot 12$

Finally, substituting the known values, we can conclude the value of $r$ .

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The cubic equation $x^3 + px^2 + qx + r = 0$, where $p$, $q$, and $r$ are integers, has roots $\alpha$, $\beta$, and $\gamma$, such that: $\alpha + \beta + \gamma = 15$ - CIE - A-Level Further Maths - Question 5 - 2015 - Paper 1

Worked Solution & Example Answer:The cubic equation $x^3 + px^2 + qx + r = 0$, where $p$, $q$, and $r$ are integers, has roots $\alpha$, $\beta$, and $\gamma$, such that: $\alpha + \beta + \gamma = 15$ - CIE - A-Level Further Maths - Question 5 - 2015 - Paper 1

Write down the value of $p$

find the value of $q$

Given that $\alpha$, $\beta$ and $\gamma$ are all real and that $\alpha \beta + \gamma = 36$, find $\alpha$

hence find the value of $r$

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