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# Janusz Asked

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 $$a = -{1\over 2},\ \alpha = {-3 \pm \sqrt 5\over 2},\ b = {-5 \pm 2\sqrt 5\over 2},\ \beta = {-7 \pm 3\sqrt 5\over 2},\ c = {-9 \pm 4\sqrt 5\over 2}.$$

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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 $$a = -{1\over 2},\ \alpha = {-3 \pm \sqrt 5\over 2},\ b = {-5 \pm 2\sqrt 5\over 2},\ \beta = {-7 \pm 3\sqrt 5\over 2},\ c = {-9 \pm 4\sqrt 5\over 2}.$$

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