Janusz Asked

In y = ax +b when are a, -b/a, b in arithmetic progression. The polynomial y = ax^2 + bx + c has roots r1 and r2. Can a, r1, b, r2 and c be in arithmetic progression?

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Stage: 5 Challenge Level:

Good solutions to this problem were received from Tyrone of Cyfarthfa High School in Merthyr Tydfil, and Koopa of Boston College in the USA.

Tyrone solved the problem by relating both polynomials to $(x+1)^2$ :

$$ \eqalign { p(x)=x^2 + 2x \Rightarrow &p &=&(x+1)^2 - 1 \\ &p+1 &=& (x+1)^2 \\ } $$

(1)

$$ \eqalign { q(x)=x^2 + x + 1 \Rightarrow &q &=&(x+1)^2 - x \\ &q+x &=& (x+1)^2 \\ } $$

(2)

$ \Rightarrow p+1=q+x $ (combining eqns (1) and (2)).

But $x=(p+1)^{1/2}-1$ (from eqn (1)). So

$$ \eqalign { \Rightarrow p+1&=&q+ ((p+1)^{1/2}-1) \\
&=&q+ (p+1)^{1/2}-1 }$$
$$ \eqalign { p-q+2&=&(p+1)^{1/2} \\
(p-q+2)^2&=&p+1}$$.

Squaring the bracket,

$$ \eqalign { &p^2-pq+2p-pq+q^2-2q+2p-2q+4=p+1 \\ &p^2 -2pq+q^2+4p-4q+4=p+1 \\ &p^2-2pq+q^2+3p-4q+3=0 } $$.

Inequality/inequalities. Manipulating algebraic expressions/formulae. Making and proving conjectures. Generalising. Patterned numbers. Other equations. Mathematical reasoning & proof. Number theory. Polynomials. Relations.

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