# Prove that a property holds for the roots of an equation, given some constraints.

For example, in this question:

The cubic expression $$ax^3 + bx^2 + cx + d$$ has a pair of roots which are reciprocals of each other. Prove that $$a^2-d^2=ac-bd$$.

I understand that the correct approach is probably to use the formulae relating coefficients and roots, but I'm not exactly sure where to go for there.

Is there are a more generic framework for solving these types of problems?

• You might want to look at Vieta's formulae/relations which relate the coefficients of a polynomial to the elementary symmetric functions of the roots - but for this kind of problem you are probably looking for a short cut. – Mark Bennet Jan 24 at 20:35
• Sorry, I meant reciprocals. – hidden-shelter Jan 24 at 20:37
• I've updated it accordingly. – hidden-shelter Jan 24 at 20:37
• Thank you for clarifying. – Mark Bennet Jan 24 at 20:38

Let the roots be $$y, z, \frac{1}{z}$$.

As you said, "use the formulae relating coefficients and roots" to show that

• $$y+z + \frac{1}{z} = -b/a$$
• $$yz + y/z + 1 = c/a$$
• $$y = -d/a$$

Finally, verify that

$$1 - (d/a)^2 = 1-y^2 = (c/a) - (b/a)(d/a).$$

Hence, $$a^2 - d^2 = ac - bd$$.

• Thanks! I'm just not sure as to how the verification leads to the proof that $a^2-d^2 = ac - bd$? – hidden-shelter Jan 24 at 20:53
• As I stated/implied, show that $1 - (d/a) ^2 = 1- y^2$, and that $1- y^2 = (c/a) - (b/a)(d/a)$, then multiply throughout by $a^2$. – Calvin Lin Jan 24 at 20:54
• But how does this relate to $a^2 - d^2$? – hidden-shelter Jan 24 at 20:56
• Oh wait, no I see it now. Many thanks! – hidden-shelter Jan 24 at 20:57
• What is $a^2 \times [ 1 - (d/a) ^2 ]$? What is $a^2 \times [ (c/a) - (b/a)(d/a) ]$? If you're stuck, please show what you've tried and why you're stuck. – Calvin Lin Jan 24 at 20:57

As a complement to @CalvinLin's answer, let us denote the polynomials $$$$\begin{array}[l]\cr P(x) = a x^3 + b x^2 + c x + d\cr Q(x) = d x^3 + c x^2 + b x + a \end{array}$$$$ Explicit computation shows that the resultant of these two polynomials is $$$$R(P, Q) = \left|\begin{matrix} a&0&0&d&0&0\cr b&a&0&c&d&0\cr c&b&a&b&c&d\cr d&c&b&a&b&c\cr 0&d&c&0&a&b\cr 0&0&d&0&0&a\cr \end{matrix}\right| = -(a^2-d^2- a c + b d)^2 P(1)P(-1)$$$$ It follows that $$P$$ and $$Q$$ have a common root if and only if $$1$$ or $$-1$$ is a root of $$P$$ or $$a^2-d^2 = a c - b d$$. Note that $$1$$ and $$-1$$ are their own reciprocals.

Here is the (python) code to prove the claim

import sympy as sp

a, b, c, d = sp.symbols(['a','b','c','d'])

M = sp.Matrix([
[a, b, c, d, 0, 0],
[0, a, b, c, d, 0],
[0, 0, a, b, c, d],
[d, c, b, a, 0, 0],
[0, d, c, b, a, 0],
[0, 0, d, c, b, a],
])
p = a**2 - d**2 - a * c + b * d
q = (a + b + c + d) * (a - b + c - d)
R = M.det()
s = R - p**2 * q
assert s.simplify() == 0

• Why not sp.factor(R)? – Marc Glisse Jan 25 at 7:58
• @MarcGlisse You are right. There is even a "resultant" in sympy. I should do more formal calculations... – Gribouillis Jan 26 at 15:26

Okay, so if $$r$$ and $$\frac 1r$$ are two roots of $$ax^3 + bx^2 + cx + d$$ then $$ax^3 + bx^2 + cx + d = a(x-r)(x-\frac 1r)(x+k)$$ for some $$k$$.

So $$ka = d$$

And $$ak -ar -a\frac 1r = d-ar -a\frac 1r = b$$.

$$-akr-ak\frac 1r+a = -dr-d\frac 1r + a=c$$

If we assume $$a\ne 0$$ then $$r+\frac 1r = \frac {d-b}a$$ and $$a = c+\frac {d-b}a\cdot d$$.

So $$a^2 =ca +d(d-b) = ca + d^2 -bd$$ and $$a^2 -d^2 =ca -db$$.

And if $$a = 0$$ then $$d=b$$. and the result follows.

(If $$a = 0; b\ne 0$$ then $$bx^2 + cx +d = b(x-r)(x-\frac 1r)= bx^2 -b(r+\frac 1r)x + b$$ so $$b=d$$ )