Using Symmetry: Additional Practice (Set 1)

Use $x^2 + y^2 = s^2 - 2p$ with $s = 6$ and $p = 7$.

$6^2 - 2(7) = 36 - 14 = 22$

Subtract the cross term $2p = 14$.

Use $x^3 + y^3 = s^3 - 3ps$ with $s = 9$ and $p = 20$.

$9^3 - 3(20)(9) = 729 - 540 = 189$

The correction term $3ps = 540$ carries the factor of $s$.

Use the template $t^2 - (\text{sum})t + (\text{product})$.

$t^2 - 6t + 4 = 0$

The sum enters with a minus sign, the product with a plus.

Recover $s$ from $s^2 = (x^2 + y^2) + 2xy$.

$s^2 = 34 + 30 = 64 \implies s = 8$

Taking the positive value gives $x + y = 8$.

Add the reciprocals as the sum over the product.

$\frac{1}{x} + \frac{1}{y} = \frac{s}{p} = \frac{5}{6}$

Use $(x - y)^2 = s^2 - 4p$ with $s = 2$ and $p = -15$.

$2^2 - 4(-15) = 4 + 60 = 64$

Subtracting a negative product adds, so the answer is $64$.

Swap $x$ and $y$ and see which is unchanged.

$x^2 + y^2 + xy \ \longrightarrow \ y^2 + x^2 + yx$

Only the first is identical after the swap; the others change sign or value.

Sum $8$ and product $12$ give $t^2 - 8t + 12 = 0$.

$(t - 2)(t - 6) = 0$

The roots are $2$ and $6$, so $\{x, y\} = \{2, 6\}$.

Use $x^2 + \frac{1}{x^2} = k^2 - 2$ with $k = 2$.

$2^2 - 2 = 4 - 2 = 2$

Here $x + \frac{1}{x} = 2$ forces $x = 1$, and indeed $1 + 1 = 2$.

Build $t^2 - st + p$ with $s = -6$ and $p = 8$.

$t^2 + 6t + 8 = (t + 2)(t + 4) = 0$

The roots are $-2$ and $-4$, so $\{x, y\} = \{-2, -4\}$.

A real square cannot be negative, so $s^2 - 4p \ge 0$.

$s^2 \ge 4p$

This is the condition for a real pair to exist.

A symmetric system fixes the unordered pair, not which variable is which.

$\{x, y\} = \{4, 9\}$

So $(x, y)$ could be $(4, 9)$ or $(9, 4)$; both satisfy the system.