Find the cubic equation with roots $\alpha, \beta$ and $\gamma$ such that
$\alpha + \beta + \gamma = 3,$
$\alpha^2 + \beta^2 + \gamma^2 = 1,$
$\alpha^2 + \beta^3 + \gamma^3 = -30,$
giving your answer in the form $x^3 + px^2 + qx + r = 0,$ where $p, q$ and $r$ are integers to be found. - CIE - A-Level Further Maths - Question 2 - 2016 - Paper 1
Question 2
Find the cubic equation with roots $\alpha, \beta$ and $\gamma$ such that
$\alpha + \beta + \gamma = 3,$
$\alpha^2 + \beta^2 + \gamma^2 = 1,$
$\alpha^2 + \beta... show full transcript
Worked Solution & Example Answer:Find the cubic equation with roots $\alpha, \beta$ and $\gamma$ such that
$\alpha + \beta + \gamma = 3,$
$\alpha^2 + \beta^2 + \gamma^2 = 1,$
$\alpha^2 + \beta^3 + \gamma^3 = -30,$
giving your answer in the form $x^3 + px^2 + qx + r = 0,$ where $p, q$ and $r$ are integers to be found. - CIE - A-Level Further Maths - Question 2 - 2016 - Paper 1
Step 1
Find Expressions for $p$, $q$, and $r$
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Answer
Using Vieta's formulas, we know:
p=−(α+β+γ)
From the equation, we have:
p=−3.
Step 2
Calculate $\alpha^2 + \beta^2 + \gamma^2$
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Answer
We use the identity:
α2+β2+γ2=(α+β+γ)2−2(αβ+βγ+γα).
Substituting known values:
α2+β2+γ2=32−2s=1,
where s=αβ+βγ+γα. Thus,
9−2s=1⇒2s=8⇒s=4.
Step 3
Calculate $\alpha^3 + \beta^3 + \gamma^3$
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Answer
We use the identity:
α3+β3+γ3=(α+β+γ)(α2+β2+γ2−αβ−βγ−γα)+3αβγ.
Substituting values gives us:
−30=3(1−4)+3r⇒−30=3(−3)+3r⇒−30=−9+3r⇒3r=−21⇒r=−7.
Step 4
Formulate the Cubic Equation
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Answer
Finally, substituting values of p, q, and r into the cubic equation:
x3−3x2+4x−7=0.
Thus, the cubic equation is:
x3−3x2+4x−7=0.
