# Finding roots of quartic polynomial using a quadratic polynomial of similar form.

Question : Let $P(x)$ be the quadratic $Ax^2 + Bx + C$. Suppose that $P(x) = x$ has unequal real roots. Show that the roots are also roots of $P(P(x)) = x$. Find a quadratic equation for the other two roots of this equation. Hence solve $(y^2 - 3y + 2)^2 - 3(y^2 - 3y + 2) + 2 - y = 0$.

The first part is easy to show, Let $a,b$ be roots of $P(x)=x$ then we use $P(a)=a$ twice to show that it is the root of given polynomial $P(P(x)) = x$.

I am having difficulty in the second part, construction of a similar polynomial whose roots are the remaining two roots. Please help.

Thank you.

## 4 Answers

Suppose that the root of $Ax^2+(B-1)x+C$ are $\alpha$ and $\beta$ . Then $\alpha+\beta=\frac{-(B-1)}{A}$ and $\alpha\beta=\frac{C}{A}$. Consider $A(Ax^2+(B)x+C)^2+B(Ax^2+(B)x+C)+C-x$. Two of its roots are known . If we use sum of roots formula and subtract the sum of known roots we get the sum of remaining two roots as $-\frac{B+1}{A}$. Similiarly products of two remaining roots is $\frac{AC+B+1}{A^2}$. Hence the quadratic equation for two other roots is $A^2x^2+(AB+A)x+AC+B+1=0$.

The special case: Since the (NB: distinct) roots of $P(y)-y$ are also both roots of $P(P(y))-y$, the first polynomial must be a divisor of the second, so it suffices to make a polynomial division: $$P(P(y))-y= y^4-6y^3+10y^2-4 y : P(y)-y = y^2-4y+2= y^2-2y$$ from which we get the supplementary roots: 0 and 2. (In addition to $2\pm \sqrt{2}$ coming from $P(y)-y$).

For the general case we have (NB: the principle being valid for any polynomial not just quadratic): $$P(P(y))-y = P(P(y))-P(y)+P(y)-y = A(P(y)^2-y^2)+ B(P(y)-y) +P(y)-y$$ or $$P(P(y))-y = (P(y)-y) (A(P(y)+y)+B+1)$$ The other equation is therefore $$A(P(y)+y) +B+1= A^2y^2 +A(1+B)y +(B+AC+1)=0$$ which in our special case yields precisely $y^2-2y=0$. The general result is valid also when the roots of $P$ are not distinct.

If there exists and $x$ such that $P(x) = x$ then $P(P(x)) = P(x) = x$

$(y^2−3y+2)^2−3(y^2−3y+2)+2-y = 0\\ (y^2−3y+2)^2−3(y^2−3y+2)+2=y$

$P(P(y)) = y\\ P(y) =y^2−3y+2$

Find $y$ such that: $y^2−3y+2=y$

$y^2 - 4y +2 = 0\\ y = 2\pm\sqrt{2}$

This gives us two roots of: $(y^2−3y+2)^2−3(y^2−3y+2)+2-y$

$2$ is also a root: $(y^2−3y+2)^2−3(y^2−3y+2)+2-y = (y-2)[(y-1)^2(y-2) −3(y-1) - 1]$

That leaves one root to find.

The product of the roots equals the constant term of the polynomial $P(P(0))-0 = P(2) = 0$

$0$ is a root.

$P(P(y)) - y = (y-2)(y-2-\sqrt{2})(y-2+\sqrt{2})y$

• Well, the OP was interested in the theoretical result: deriving the equation in general, for arbitrary coefficients. – Alex M. Sep 2 '16 at 18:55
• @AlexM. I thought it was rather unclear what had been requested. – Doug M Sep 2 '16 at 19:00

Let $r_1, r_2$ be the "first" roots and let $x_1, x_2$ be the "other" roots. Expanding the powers in the equation $P(P(x)) = x$ you get

$$A (Ax^2 + Bx + C)^2 + B (Ax^2 + Bx + C) + C = x \iff \\ A^3 x^4 + 2 A^2 B x^3 + (A B^2 + 2 A^2 C + AB) x^2 + (2ABC + B^2 - 1) x + (A C^2 + BC + C) = 0 .$$

By Vieta's relations we have that

$$r_1 + r_2 + x_1 + x_2 = - \frac {2 A^2 B} {A^3} = - \frac {2B} A, \quad r_1 r_2 x_1 x_2 = \frac {A C^2 + BC + C} {A^3} .$$

On the other hand, $r_1, r_2$ satisfy the equation $P(P(r)) = r$, which means that

$$A r^2 + (B-1) r + C = 0 ,$$

whence again by Vieta's relations we obtain

$$r_1 + r_2 = - \frac {B-1} A, \quad r_1 r_2 = \frac C A .$$

Combining the results obtained from the two uses of Vieta's relations, we get that

$$x_1 + x_2 = - \frac {2B} A + \frac {B-1} A = - \frac {B+1} A \\ x_1 x_2 = \frac {A C^2 + BC + C} {A^3} \frac A C = \frac {AC + B + 1} {A^2} .$$

Since we now know the sum and the product of $x_1, x_2$, it is easy to write the desired equation as

$$x^2 - (x_1 + x_2) x + x_1 x_2 = 0 \iff x^2 + \frac {B+1} A x + \frac {AC + B + 1} {A^2} = 0 \iff \\ A^2 x^2 + A (B+1) x + (AC + B + 1) = 0 .$$

Interestingly, I do not understand the assumption that $P(x) = x$ has unequal real roots, it seems that I haven't used it anywhere.

Solving that concrete equation is easy now, if you take $P(y) = y^2 - 3y + 2$. Then

$$A = 1, \ B = -3, C = 2$$

so the "other" equation is

$$y^2 - 2y = 0 ,$$

with roots $0$ and $2$. Since $P(y) = y$ has roots $2 + \sqrt 2$ and $2 - \sqrt 2$, that 4th degree equation has roots $\{0, 2 - \sqrt 2, 2, 2 + \sqrt 2\}$.

• x=-7 is not the root of the equation – sn24 Sep 2 '16 at 18:40
• @sn24: Fixed, thank you. – Alex M. Sep 2 '16 at 18:52