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Real(ly) Numbers

If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?

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?

Polynomial Relations

Given any two polynomials in a single variable it is always possible to eliminate the variable and obtain a formula showing the relationship between the two polynomials. Try this one.

More Polynomial Equations

Age 16 to 18
Challenge Level

Jim sent in this solution, using the ideas from our hints.

$a(n)=1+2+\ldots+n=\frac{n(n+1)}{2}$
$b(n)=1^2+2^2+\ldots+n^2=\frac{n(n+1)(2n+1)}{6}$
$c(n)=1^3+2^3+\ldots+n^3=\frac{n^2(n+1)^2}{4}$
It's obvious that $c=a^2$, from this.

Also, $2a=n^2+n$, so, solving the quadratic (and using the fact that $n> 0$), we get $n=\frac{-1+\sqrt{1+8a}}{2}$.

Now substitute this for $n$ in $b$, to get $b=\frac{n(n+1)(2n+1)}{6}=\frac{a}{3}\times(2n+1)=\frac{a}{3}\times\sqrt{1+ 8a}$. So $3b=a\sqrt{1+8a}$, so $9b^2=a^2+8a^3$.

Now we can combine these two expressions: $9b^2=c+8c\sqrt{c}$, so $8c\sqrt{c}= 9b^2-c$, so $64c^3=81b^4-18b^2c+c^2$.

(It's easy to check that the expressions above in terms of $n$ do work in this!)