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We start with the linear polynomial $y = -3x + 9$. The
$x$-coefficient is $-3$, the root is $3$, the intercept is $9$, and
these numbers are in arithmetic progression. Suppose now that the
same is true of the polynomial $y = ax + b$, where $a\neq 0$. In
this case, the numbers $a$, $-b/a$ and $b$ must be in arithmetic
progression. Does this enable you to find a relationship between
$a$ and $b$ and hence to find a condition giving infinitely many
polynomials with this property? Now consider the quadratic
polynomial $$y = ax^2 + bx + c$$ with roots $r_1$ and $r_2$, and
suppose that the numbers $a$, $r_1$, $b$, $r_2$ and $c$ are in
arithmetic progression. You may find it easiest here to use the sum
and product of the roots, namely $-b/a$ and $c/a$ and the common
difference, and to avoid using the quadratic formula.
Andaleeb Ahmed of Woodhouse
Sixth Form College, London has provided a solution to this
problem.
We started with the linear polynomial $y = -3x + 9$. The
$x$-coefficient is $-3$, the root is $3$, the intercept is $9$, and
these numbers are in arithmetic progression. Suppose now that the
same is true of the polynomial $y=ax+b$, where $a\neq 0$. In this
case, the numbers $a$, $-b/a$ and $b$ must be in arithmetic
progression.
Then: $${-b\over a} - a = b +{b\over a}$$ $$-b - a^2 = ab + b$$ $$
a^2 + ba + 2b = 0$$ We can use this to express $a$ in terms of $b$
or $b$ in terms of $a$: $$a = {{-b\pm \sqrt{b^2 - 8b}}\over 2}\quad
\quad (b\geq 8 \ {\rm or}\ b < 0).$$ $$b={-a^2\over (a+2)}\quad
\quad (a\neq -2).$$ We can randomly choose $b$ (such that $b\geq 8$
or $b < 0$) and use the formula to find the corresponding values
of $a$, or alternatively, we could choose any value of $a$ except
$a = -2$ and find the corresponding value of $b$ from the given
formula.
The linear polynomial is $$y = ax - \left({a^2\over a+2}\right).$$
It is easily checked that this polynomial does have the required
property for any value of $a$ and clearly there are infinitely many
such polynomials.
Now consider $y = ax^2 + bx +c$, where $a, \alpha, b, \beta,c$ are
in arithmetic progression. Using the sum of the roots: $$\beta +
\alpha = -{b\over a} \quad (1).$$ Also, since they are in
arithmetic progression: $$\beta -\alpha = b - a \quad (2).$$ Adding
the two equations we get: $$\beta = {-b + ab - a^2\over 2a}.$$
Subtracting the two equations we get: $$\alpha = {-b - ab +
a^2\over 2a}.$$ Since $a, \alpha, b, \beta, c$ are in arithmetic
progression: $$\alpha - a = b - \alpha$$ $$\Rightarrow 2\alpha = a
+ b $$ $$\Rightarrow {-b - ab + a^2\over a} = a + b.$$ By solving
this equation , we get $b = 0$ or $a = -1/2$.
Case
1: $b = 0$
Since $\alpha - a = b - \alpha$ we have $\alpha = a/2$ and
similarly $\beta = c/2.$ Now it can be said that $a, {a\over 2}, 0,
{c\over 2}, c$ are in arithmetic progression, hence $c = -a$.
Thus the equation $y = ax^2 + bx + c$ becomes $y = ax^2 - a$ with
$\alpha = {a\over 2}$ and $\beta = -{a\over 2}$. Hence $$0 = a\big(
({a\over 2})^2 - 1\big) = a\big( ({-a\over 2})^2 - 1\big).$$ From
these equations we get $a = \pm 2$ or 0. Thus when $b = 0$, we have
the following equations that satisfy the above property: $y = 2x^2
- 2$ and $y = -2x^2 + 2.$
Case
2: $a = -1/2$
Here $\alpha = {a + b\over 2} = {2b - 1\over 4}.$ Since $\beta -
\alpha = b - a$ $$\beta = {2b - 1\over 4} + b + {1\over 2} = {6b +
1\over 4}.$$ Now it can be said that $-{1\over 2},\ {2b - 1\over
4},\ b,\ {6b + 1\over 4},\ c$ are in arithmetic progression. So
$${6b + 1\over 4} - {2b - 1\over 4} = c - b$$ and thus $c = {4b +
1\over 2}.$ Thus $y = ax^2 + bx + c$ becomes $y = -{1\over 2}x^2 +
bx + {4b+1\over 2}.$ As we know $y=0$ when $x=\alpha$ this gives
$$-{1\over 2}\big({2b-1\over 4}\big)^2 + b\big({2b - 1\over 4}\big)
+ {4b+1\over 2} = 0.$$ Simplifying and rearranging the equation we
get: $$12b^2 + 60b + 15 = 0.$$ Thus $$b = {-5 \pm 2\sqrt 5\over
2},$$ and $$c = {4b + 1\over 2} = {-9 \pm 4\sqrt 5\over 2},$$ and
so, when $a = -{1\over 2}$, we have quadratic polynomials with the
required property for these values of $b$ and $c$.
It can now be said that there are a finite number of quadratic
polynomials $y = ax^2 + bx + c$ for which $a,\ \alpha,\ b,\ \beta,$
and $c$ are in arithmetic progression, and these occur when $a =
\alpha = b = \beta = c = 0$ (trivially) or when $$a = \pm 2,\
\alpha = \pm 1,\ b = 0,\ \beta = \mp 1, \ c = \mp 2$$ or when $$a =
-{1\over 2},\ \alpha = {-3 \pm \sqrt 5\over 2},\ b = {-5 \pm 2\sqrt
5\over 2},\ \beta = {-7 \pm 3\sqrt 5\over 2},\ c = {-9 \pm 4\sqrt
5\over 2}.$$