Algebra & Inequalities | The Maths Tailor — UK Methods
Algebra & Inequalities
7 methods · 6 questions
When a polynomial equation has symmetric (palindromic) coefficients, its non-zero roots come in reciprocal pairs z,z−1, so divide through by the middle power and substitute w=z+z−1 (or w=z−z−1 when the symmetry is anti-palindromic) to fold the equation onto one of half the degree.
Trigger: coefficients read the same (or with alternating sign) forwards and backwards, or you can verify that k being a root forces k−1 to be a root.
Instances: (i) Even palindromic quartic z4+az3+bz2+az+1: divide by z2 and set w=z+z−1 to get a quadratic in w. (ii) Anti-palindromic case (signs alternate): set w=z−z−1 instead, so the cross terms cancel. (iii) Higher even degree (degree 8): the same divide-and-substitute collapses it to a quartic in w, then iterate. (iv) Use the structural fact alone: prove k root ⟹k−1 root, which pins down how repeated or distinct roots can be distributed.
Linked questions (1)
STEP 2 2018 — Q1
Show that, if k is a root of the quartic equation x4+ax3+bx2+ax+1=0,(∗) then k−1 is a root.
You are now given that a and b in (∗) are both real and are such that the roots are all real.
Write down all the values of a and b for which (∗) has only one distinct root.
Given that (∗) has exactly three distinct roots, show that either b=2a−2 or \text{b=−2a−2}.
Solve (∗) in the case b=2a−2, giving your solutions in terms of a.
Given that a and b are both real and that the roots of (∗) are all real, find necessary and sufficient conditions, in terms of a and b, for (∗) to have exactly three distinct real roots.
Voir la correctionMasquer la correction
The key observation is that the quartic x4+ax3+bx2+ax+1=0 is palindromic: its coefficients read the same forwards and backwards. If k is a root and k=0, substitute x=k−1:k−4+ak−3+bk−2+ak−1+1=0.Multiplying through by k4 gives 1+ak+bk2+ak3+k4=0, which is precisely the original equation evaluated at k. Since k is a root, this expression equals zero, so k−1 is also a root. (Note k=0 cannot be a root since the constant term is 1.)
Part (i). For only one distinct root, that root r must satisfy r=r−1, so r=±1.
- If r=1: the quartic is (x−1)4=x4−4x3+6x2−4x+1, giving a=−4, b=6. - If r=−1: the quartic is (x+1)4=x4+4x3+6x2+4x+1, giving a=4, b=6.
Part (ii). With exactly three distinct roots, one root must be repeated. The palindrome symmetry pairs roots as {k,k−1}, so a repeated root must satisfy k=k−1, i.e. k=±1.
If x=1 is a repeated root, substitute x=1 into (∗):1+a+b+a+1=0⟹b=−2a−2.If x=−1 is a repeated root, substitute x=−1:1−a+b−a+1=0⟹b=2a−2.Thus either b=2a−2 or b=−2a−2.
Part (iii). Take b=2a−2 (repeated root at x=−1). Factor out (x+1)2:(x+1)2(x2+cx+1)for some c. Expanding: (x2+2x+1)(x2+cx+1)=x4+(c+2)x3+(3+2c)x2+(c+2)x+1.
Matching the coefficient of x3: c+2=a, so c=a−2.
The remaining roots come from x2+(a−2)x+1=0:x=2−(a−2)±(a−2)2−4=2(2−a)±a2−4a.So the roots of (∗) in this case are x=−1 (repeated) and x=2(2−a)±a2−4a.
Necessary and sufficient conditions for exactly three distinct real roots.
We need the quadratic factor to have two real roots, neither of which equals the repeated root.
Case b=2a−2 (repeated root −1): Require discriminant >0:(a−2)2−4>0⟹a2−4a>0⟹a(a−4)>0,so a<0 or a>4. Check the quadratic roots are not −1: substituting x=−1 gives 1−(a−2)+1=4−a, which is zero only if a=4. Since a=4 is excluded by the strict inequality, this never occurs.
Case b=−2a−2 (repeated root 1): By symmetry (a↦−a), the condition is a>0 or a<−4, equivalently a(a+4)>0.
Therefore the necessary and sufficient conditions are:b=2a−2 with a<0 or a>4,or b=−2a−2 with a>0 or a<−4.
