Sum and Product of Roots: Additional Practice (Set 2)

Only symmetric combinations of the roots are fixed by the coefficients. Swapping $r_1$ and $r_2$ leaves $r_1 + r_2$, $r_1 r_2$, and $r_1^2 + r_2^2$ unchanged, but it flips the sign of $r_1 - r_2$.

$b, c \text{ fix } (r_1 - r_2)^2 = \frac{\Delta}{a^2}, \text{ but not } r_1 - r_2$

The coefficients cannot decide which root is larger, so $r_1 - r_2$ is not determined.

The roots of $2x^2 - 7x + 3$ are $3$ and $\tfrac{1}{2}$, so the reciprocals are $\tfrac{1}{3}$ and $2$, with sum $\tfrac{7}{3}$ and product $\tfrac{2}{3}$. The monic quadratic is $x^2 - \tfrac{7}{3}x + \tfrac{2}{3}$; multiply by $3$.

$3\left(x^2 - \tfrac{7}{3}x + \tfrac{2}{3}\right) = 3x^2 - 7x + 2$

Reversing the original coefficients gives the reciprocal-root quadratic.

Use $r_1^3 + r_2^3 = (r_1 + r_2)^3 - 3 r_1 r_2 (r_1 + r_2)$ with sum $3$ and product $-2$.

$r_1^3 + r_2^3 = 3^3 - 3(-2)(3) = 27 + 18 = 45$

The negative product raises the total.

A repeated root requires $\Delta = k^2 - 4(16) = 0$.

$k^2 = 64 \;\Rightarrow\; k = \pm 8$

Both signs give a real double root.

The roots of $x^2 - 5x + 6$ have sum $5$ and product $6$. The new roots $r_1 + 2$ and $r_2 + 2$ have sum $(r_1 + r_2) + 4 = 9$ and product $r_1 r_2 + 2(r_1 + r_2) + 4 = 6 + 10 + 4 = 20$.

$x^2 - 9x + 20$

The new sum and product feed straight into the monic template.

Use $\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2}$. The sum is $\frac{5}{3}$ and the product is $\frac{2}{3}$.

$\frac{1}{r_1} + \frac{1}{r_2} = \frac{5/3}{2/3} = \frac{5}{2}$

The threes cancel in the ratio.

The roots multiply to the constant $-15$, so the other root is $\frac{-15}{-5} = 3$. The roots then sum to $-5 + 3 = -2$, and the sum equals $-b$.

$-b = -2 \;\Rightarrow\; b = 2$

The other root is $3$ and $b = 2$.

Use $(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4 r_1 r_2$ with sum $9$ and product $14$.

$(r_1 - r_2)^2 = 9^2 - 4(14) = 81 - 56 = 25$

The roots $2$ and $7$ differ by $5$, and $5^2 = 25$.

The condition $(r_1 - r_2)^2 = 16$ becomes $(r_1 + r_2)^2 - 4 r_1 r_2 = 36 - 4c = 16$.

$36 - 4c = 16 \;\Rightarrow\; c = 5$

The discriminant $36 - 20 = 16 \ge 0$, so the roots are real.

Factor out $r_1 r_2$: the expression is $r_1 r_2 (r_1 + r_2)$, with sum $7$ and product $10$.

$r_1^2 r_2 + r_1 r_2^2 = 10 \cdot 7 = 70$

It reduces to the product times the sum.

Expand the square of the sum: $(r_1 + r_2)^2 = r_1^2 + 2 r_1 r_2 + r_2^2$. Subtracting $2 r_1 r_2$ removes the cross term.

$(r_1 + r_2)^2 - 2 r_1 r_2 = r_1^2 + r_2^2$

Note that subtracting $4 r_1 r_2$ instead would give $(r_1 - r_2)^2$, a different quantity.

Use the template $x^2 - (\text{sum})x + (\text{product})$ with sum $-3$ and product $-4$.

$x^2 - (-3)x + (-4) = x^2 + 3x - 4$

The negative sum becomes a positive middle coefficient after the sign flip.