The roots of the quartic equation $x^4 + 4x^3 + 2x^2 - 4x + 1 = 0$ are $\alpha, \beta, \gamma$ and $\delta$ - CIE - A-Level Further Maths - Question 11 - 2014 - Paper 1
Question 11
The roots of the quartic equation $x^4 + 4x^3 + 2x^2 - 4x + 1 = 0$ are $\alpha, \beta, \gamma$ and $\delta$. Find the values of:
(i) $\alpha + \beta + \gamma + \del... show full transcript
Worked Solution & Example Answer:The roots of the quartic equation $x^4 + 4x^3 + 2x^2 - 4x + 1 = 0$ are $\alpha, \beta, \gamma$ and $\delta$ - CIE - A-Level Further Maths - Question 11 - 2014 - Paper 1
Step 1
(i) $\alpha + \beta + \gamma + \delta$
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Using Vieta's formulas, the sum of the roots of the polynomial is derived from the coefficient of the x3 term. Therefore,
α+β+γ+δ=−14=4.
Step 2
(ii) $\alpha^2 + \beta^2 + \gamma^2 + \delta^2$
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Using the relation:
α2+β2+γ2+δ2=(α+β+γ+δ)2−2(αβ+αγ+αδ+βγ+βδ+γδ)
The sum of the products of the roots taken two at a time is given by 12=2. Hence,
Sign up now to view full answer, or log in if you already have an account!
Answer
Using the property of reciprocals of roots:
α1+β1+γ1+δ1=αβγδαβγ+αβδ+αγδ+βγδ
Thus,
=1(−4)=−4.
Step 4
(iv) Using the substitution $y = x + 1$, find a quartic equation in $y$.
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Substituting y=x+1 implies x=y−1. Hence:
4x^3 = 4(y-1)^3,
2x^2 = 2(y-1)^2,
-4x = -4(y-1)$$
Expanding and simplifying leads to:
$$y^4 - 6y^3 + 8y^2 - 6y + 1 = 0.$$
To solve for the roots of this equation, use numerical methods or the quadratic formula, where
$$y^2 - 2y - 1 = 0$$
yielding roots \(y = 1 \pm \sqrt{2}.
The roots of the original equation can be traced back from here.
