Vieta's Formulas: Additional Practice (Set 2)

Divide by the leading coefficient in both symmetric sums: $e_1 = -\frac{-4}{2} = 2$ and $e_2 = \frac{3}{2}$.

Now apply the identity:

$r_1^2 + r_2^2 + r_3^2 = e_1^2 - 2e_2 = 4 - 2 \cdot \frac{3}{2} = 4 - 3 = 1$

The three wrong choices are the three ways to fumble that last line: adding the cross terms instead of subtracting them gives $4 + 3 = 7$, dropping them altogether leaves $e_1^2 = 4$, and subtracting in the wrong order gives $3 - 4 = -1$.

The product of all the roots is $e_5 = (-1)^5\frac{a_0}{a_5}$, so the odd degree flips the sign, and the leading coefficient still divides.

$e_5 = (-1)^5\,\frac{-8}{2} = -(-4) = 4$

Two minus signs are in play, one from the parity and one from the constant term, and they cancel.

Expand the product and watch the symmetric sums appear on their own.

$(1 + r_1)(1 + r_2)(1 + r_3) = 1 + (r_1 + r_2 + r_3) + (r_1r_2 + r_1r_3 + r_2r_3) + r_1r_2r_3 = 1 + e_1 + e_2 + e_3$

Vieta gives $e_1 = 2$, $e_2 = -5$, and $e_3 = -6$, so the value is $1 + 2 - 5 - 6 = -8$.

The roots are $1$, $3$, $-2$, and $(2)(4)(-1) = -8$ confirms it.

Read the symmetric sums off the coefficients: $e_1 = 5$ and $e_2 = 6$.

$r_1^2 + r_2^2 + r_3^2 = e_1^2 - 2e_2 = 25 - 12 = 13$

This cubic has no rational roots, so there was never any prospect of finding them by hand, and none was needed.

The product of all three roots is $-\frac{d}{a} = -\frac{-24}{1} = 24$.

That product is the two known-to-multiply roots times the third, so divide:

$r_3 = \frac{24}{6} = 4$

The roots are $2$, $3$, $4$, and the pair that multiplies to $6$ is $2$ and $3$.

The coefficient of $x^2$ in a monic quartic is $+e_2$, the sum of the products over all six pairs of entries in the list. Both copies of each repeated root take part.

$e_2 = (2)(2) + (2)(-1) + (2)(-1) + (2)(-1) + (2)(-1) + (-1)(-1) = 4 - 2 - 2 - 2 - 2 + 1 = -3$

Expanding confirms it: $(x - 2)^2(x + 1)^2 = x^4 - 2x^3 - 3x^2 + 4x + 4$.

Over the common denominator $r_1r_2r_3$, the sum of reciprocals is $\frac{e_2}{e_3}$.

Vieta, dividing by $a = 2$ each time, gives $e_2 = \frac{-11}{2}$ and $e_3 = -\frac{6}{2} = -3$.

$\frac{e_2}{e_3} = \frac{-11/2}{-3} = \frac{11}{6}$

The two minus signs cancel. The roots are $3$, $-2$, $\frac{1}{2}$, and $\frac{1}{3} - \frac{1}{2} + 2 = \frac{11}{6}$.

The constant term of a monic quartic is $+e_4$, because $(-1)^4 = +1$, so there is no sign flip to apply.

$e_4 = (2)(2)(-1)(-1) = 4$

The two negative roots multiply to a positive, and the quartic is $x^4 - 2x^3 - 3x^2 + 4x + 4$.

Put the two fractions over the common denominator $r_1r_2$, which turns the question into one about the sum of squares.

$\frac{r_1}{r_2} + \frac{r_2}{r_1} = \frac{r_1^2 + r_2^2}{r_1r_2} = \frac{e_1^2 - 2e_2}{e_2}$

Vieta gives $e_1 = 5$ and $e_2 = 3$, so the value is $\frac{25 - 6}{3} = \frac{19}{3}$. The roots are irrational and never appear.

The product of all three roots is $-\frac{d}{a} = -\frac{-36}{1} = 36$.

That total is the given pair's product times the remaining root:

$r_3 = \frac{36}{6} = 6$

The roots are $2$, $3$, $6$, and the pair multiplying to $6$ is $2$ and $3$.

Name the three pairwise products $A = r_1r_2$, $B = r_1r_3$, $C = r_2r_3$. The question asks for $A^2 + B^2 + C^2$, and the same squaring trick applies to them.

Their sum is $A + B + C = e_2 = 14$, and their pairwise products satisfy $AB + AC + BC = e_1e_3 = 7 \cdot 8 = 56$, since for instance $AB = r_1^2r_2r_3 = r_1e_3$.

$A^2 + B^2 + C^2 = (A + B + C)^2 - 2(AB + AC + BC) = 196 - 112 = 84$

The roots are $1$, $2$, $4$, so the pairwise products are $2$, $4$, $8$, and $4 + 16 + 64 = 84$.

Read the four symmetric sums off the coefficients, flipping the sign whenever $k$ is odd.

They are $e_1 = -3$, $e_2 = -2$, $e_3 = -1$, and $e_4 = -1$.

$e_1 + e_2 + e_3 + e_4 = -3 - 2 - 1 - 1 = -7$

A check: $1 + e_1 + e_2 + e_3 + e_4$ equals $(1 + r_1)(1 + r_2)(1 + r_3)(1 + r_4) = p(-1) = -6$, and $-6 - 1 = -7$ as found.