Sums and Products of Roots: Practice

For $ax^2 + bx + c$ the roots add to $-\frac{b}{a}$. Here $a = 2$ and $b = -7$.

$r + s = -\frac{b}{a} = -\frac{-7}{2} = \frac{7}{2}$

So the sum is $\tfrac{7}{2}$. The value $-\tfrac{7}{2}$ forgets that $b = -7$ already carries a minus.

For $ax^2 + bx + c$ the roots multiply to $\frac{c}{a}$. Here $a = 2$ and $c = 3$.

$rs = \frac{c}{a} = \frac{3}{2}$

So the product is $\tfrac{3}{2}$. Dividing by $a$ is what separates this from the monic case.

Here $a = 3$ and $b = 5$, and the roots add to $-\frac{b}{a}$.

$r + s = -\frac{b}{a} = -\frac{5}{3}$

So the sum is $-\tfrac{5}{3}$. The value $\tfrac{5}{3}$ drops the minus sign.

Here $a = 3$ and $c = -2$, and the roots multiply to $\frac{c}{a}$.

$rs = \frac{c}{a} = \frac{-2}{3} = -\frac{2}{3}$

So the product is $-\tfrac{2}{3}$. The product keeps the sign of $c$; only the sum gets a minus.

The roots sum to $\tfrac{1}{2} + 3 = \tfrac{7}{2}$ and multiply to $\tfrac{1}{2} \cdot 3 = \tfrac{3}{2}$, so the monic quadratic is $x^2 - \tfrac{7}{2}x + \tfrac{3}{2}$. Multiply through by $2$ to clear fractions.

$2\left(x^2 - \tfrac{7}{2}x + \tfrac{3}{2}\right) = 2x^2 - 7x + 3 = 0$

This factors as $(2x - 1)(x - 3)$, confirming the roots.

The roots add to $-\frac{b}{a} = \frac{7}{2}$. Subtract the known root from that total.

$s = \frac{7}{2} - 3 = \frac{1}{2}$

So the other root is $\tfrac{1}{2}$. The product agrees, since $\frac{c}{a} = \tfrac{3}{2}$ and $\tfrac{3}{2} \div 3 = \tfrac{1}{2}$.

The sum is $(2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$, and the product is a difference of squares.

$rs = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$

Using $x^2 - (r + s)x + rs$ gives $x^2 - 4x + 1 = 0$.

The constant $c$ equals the product of the roots.

$c = rs = (-2)(7) = -14$

So $c = -14$. The value $-5$ is the sum of the roots, which relates to $b$, not $c$.

Here $a = 5$ and $b = 3$, and the roots add to $-\frac{b}{a}$.

$r + s = -\frac{3}{5}$

So the sum is $-\tfrac{3}{5}$. The value $\tfrac{3}{5}$ drops the minus sign.

In standard form the equation is $x^2 - 6x + 5 = 0$, so $c = 5$.

$rs = c = 5$

So the product is $5$. Rearrange to standard form first, then read $c$.

Here $a = 2$, $b = 3$, and $c = -5$. Use $r + s = -\frac{b}{a}$ and $rs = \frac{c}{a}$.

$r + s = -\frac{3}{2}, \qquad rs = -\frac{5}{2}$

So the sum is $-\tfrac{3}{2}$ and the product is $-\tfrac{5}{2}$.

The other root is the total sum minus the known root.

$s = 10 - (-3) = 13$

So the other root is $13$. Subtracting a negative adds, which is easy to slip on.