Quadratic Harmony
Congratulations John from State College Area High School, Pennsylvania, USA, Andrei from School 205, Bucharest, Romania and Marcos from Cyprus on your excellent solutions to Quadratic Harmony.
Say that $x^2-a x+b=0$ has roots $\alpha$, $\beta$. Then $\alpha+\beta=a$ and $\alpha\beta=b$. Without loss of generality, $a\geq b$.
Case 1: $a=b$. Then $x^2-a x+a =0$. So $\alpha+\beta=\alpha\beta$, so $\alpha\beta-\alpha-\beta=0$, so $(\alpha-1)(\beta-1)=1$. Since $\alpha$ and $\beta$ are natural numbers, we must have $\alpha-1=1$ and $\beta-1=1$, so $\alpha=2=\beta$, so $a=4=b$.
Case 2: $a> b$. Then $\alpha+\beta> \alpha\beta$, so $(\alpha-1)(\beta-1)< 1$, so $(\alpha-1)(\beta-1)=0$, so $\alpha=1 or \beta=1$. Without loss of generality, $\alpha=1$, so $b=\beta$ and $a=\beta+1=b+1$. So the quadratics are $x^2-(b+1)x+b$ and $x^2-b x+b+1$. The first of these has roots 1 and $b$, as we expected. So we just need the second one to have natural number roots. So certainly $b^2-4b-4$ (the discriminant) is a square, say $b^2-4b-4=X^2$. Then $(b-2-X)(b-2+X)=8$. We can quickly check that we can't have 8, 1 as this gives a value of $b$ that isn't an integer. So we have $b-2+X=4$, $b-2-X=2$, so $b=5$. Now we are trying to solve $x^2-6x+5=0$ and $x^2-5x+6=0$, and these are clearly both soluble in positive integers.
So, to summarise, the only possible values are $a=4$, $b=4$, and $a=5$, $b=6$ (or obviously $a$ and $b$ reversed).
Why do this problem?
The activity involves using the sum and product of the roots of a quadratic equation, the relationship between the coefficients and the roots, the discriminant of a quadratic equation, solving simple quadratic equations and checking the solutions by substitution in the original equation. It gives practice and re-inforcement of these ideas without being tedious.
The non-standard nature of the problem encourages learners to think for themselves so developing reasoning and problem solving skills which are transferable to other situations. It is also necessary to work systematically through all possible cases to find all solutions.
Possible approach
Suggest the learners try to find some solutions for themselves, initially by trial and error, and to check their solutions by substitution in the original equations. Then ask how they might make sure they have found all the solutions.
It is sometimes useful to list what you know and then to try to apply that knowledge. In this case, as each of the two equations have coefficients that occur in the other equation, and the same roots, solving the problem must involve the relationship between the cofficients and the sum and products of the roots.
This problem can be used to motivate the neeed to know and use this relationship or as re-inforcement of the ideas or as a revision exercise.
Key questions
What are the roots of the quadratic equation $x^2 + px + q = (x - \alpha)(x - \beta) = 0$?
What is the relationship between the sum of the roots and the cofficients in the equation?
What is the relationship between the productof the roots and the cofficients in the equation?
Possible extension
The next step is to explore the relationship between the graph of a quadratic function and the roots of the corresponding quadratic equation.
Possible support
Power Quady is another non-standard problem on quadtatic equations which calls for some logical reasoning to account for all possible cases and gives practice in algebraic manipulation and in solving quadratic equations.