Using Symmetry: Additional Practice (Set 2)

The substitution $u = x + \frac{1}{x}$ gives $2u^2 - 9u + 10 = 0$, so $u = \frac{5}{2}$ or $u = 2$, leading to $x = \frac{1}{2}, 2, 1$.

$x = 2 \ \text{is a root}$

Among the choices only $2$ appears in the solution set.

Build $t^2 - st + p$ with $s = 4$ and $p = -5$.

$t^2 - 4t - 5 = (t - 5)(t + 1) = 0$

The roots are $5$ and $-1$, so $\{x, y\} = \{5, -1\}$.

First $x^2 + y^2 = s^2 - 2p = 9 - 2 = 7$. Then use $x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2$.

$7^2 - 2(1)^2 = 49 - 2 = 47$

Palindromic (mirror-coefficient) equations satisfy: if $r$ is a root, so is $\frac{1}{r}$.

$r \ \text{and} \ \frac{1}{r}$

So the roots pair up as reciprocals.

Here $s^2 = 36 = 4p$, the boundary case, so the two values coincide.

$t^2 - 6t + 9 = (t - 3)^2 = 0$

Thus $x = y = 3$, a single repeated value.

Use $x^2 + y^2 = s^2 - 2p$ with $s = 7$ and $p = 12$.

$7^2 - 2(12) = 49 - 24 = 25$

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

$s^2 = 29 + 20 = 49 \implies s = 7$

Taking the positive value gives $x + y = 7$.

Palindromic equations have roots in reciprocal pairs.

$r = 5 \implies \frac{1}{r} = \frac{1}{5}$

So $\frac{1}{5}$ must also be a root.

Use $x^2 + y^2 = s^2 - 2p$ with $s = 1$ and $p = -6$.

$1 - 2(-6) = 1 + 12 = 13$

Subtracting a negative product adds.

Compute each piece: $x^2 + \frac{1}{x^2} = 4^2 - 2 = 14$ and $x^3 + \frac{1}{x^3} = 4^3 - 3(4) = 52$.

$14 + 52 = 66$

The axis is $x = -\frac{b}{2a}$ with $a = 2$ and $b = 8$.

$x = -\frac{8}{2(2)} = -2$

The two roots sit mirrored about $x = -2$.

Every symmetric polynomial in two variables reduces to the sum and product.

$x^2 + y^2 = s^2 - 2p, \qquad x^3 + y^3 = s^3 - 3ps$

So $s$ and $p$ are enough to express any of them.