Sums and Products of Roots: Additional Practice (Set 2)

Here $a = 6$ and $c = -4$, and the roots multiply to $\frac{c}{a}$. Simplify the fraction.

$rs = \frac{-4}{6} = -\frac{2}{3}$

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

The roots add to $-b = 6$. Subtract the known root from that total.

$s = 6 - (3 - \sqrt{5}) = 3 + \sqrt{5}$

So the other root is $3 + \sqrt{5}$. The product checks it, since $(3 - \sqrt{5})(3 + \sqrt{5}) = 9 - 5 = 4 = c$.

The sum is $(3 + \sqrt{2}) + (3 - \sqrt{2}) = 6$, and the product is a difference of squares.

$rs = (3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7$

Using $x^2 - (r + s)x + rs$ gives $x^2 - 6x + 7 = 0$.

Move everything to one side to reach standard form: $3x^2 - 11x + 12 = 0$. Now $a = 3$ and $b = -11$.

$r + s = -\frac{b}{a} = \frac{11}{3}$

So the sum is $\tfrac{11}{3}$. Rearrange to standard form before reading the coefficients.

The roots sum to $-\tfrac{1}{3} + 2 = \tfrac{5}{3}$ and multiply to $-\tfrac{1}{3} \cdot 2 = -\tfrac{2}{3}$, so the monic form is $x^2 - \tfrac{5}{3}x - \tfrac{2}{3}$. Multiply through by $3$.

$3\left(x^2 - \tfrac{5}{3}x - \tfrac{2}{3}\right) = 3x^2 - 5x - 2 = 0$

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

Here $a = 6$, $b = 7$, and $c = -3$. Use $r + s = -\frac{b}{a}$ and $rs = \frac{c}{a}$.

$r + s = -\frac{7}{6}, \qquad rs = \frac{-3}{6} = -\frac{1}{2}$

So the sum is $-\tfrac{7}{6}$ and the product is $-\tfrac{1}{2}$.

In standard form the equation is $2x^2 - 5x - 12 = 0$, so $a = 2$ and $c = -12$.

$rs = \frac{c}{a} = \frac{-12}{2} = -6$

So the product is $-6$.

The product of the roots is $c$. If one root is $0$, the product is $0$ no matter the other root.

$c = rs = 0 \cdot s = 0$

So $c = 0$. For example $x^2 - 5x = x(x - 5)$ has a root at $0$ and no constant term.

The product of the roots is $\frac{c}{a} = \frac{-6}{3} = -2$. Divide by the known root to get the other.

$s = \frac{rs}{r} = \frac{-2}{3} = -\frac{2}{3}$

So the other root is $-\tfrac{2}{3}$. The product route avoids needing the unknown $b$.

The roots must sum to $-\frac{b}{a} = \frac{1}{6}$ and multiply to $\frac{c}{a} = -\frac{1}{3}$. Test each factorization's roots.

$(3x - 2)(2x + 1): \quad \tfrac{2}{3} + \left(-\tfrac{1}{2}\right) = \tfrac{1}{6}, \quad \tfrac{2}{3} \cdot \left(-\tfrac{1}{2}\right) = -\tfrac{1}{3}$

Only $(3x - 2)(2x + 1)$ passes both tests. For instance $(3x + 2)(2x - 1)$ has roots $-\tfrac{2}{3}$ and $\tfrac{1}{2}$, summing to $-\tfrac{1}{6}$.

The sum of the roots is $-b$. Set it equal to $-7$.

$-b = -7 \ \Rightarrow \ b = 7$

So $b = 7$. The sum being negative does not make $b$ negative here; the minus signs cancel.

The roots must sum to $-\frac{b}{a} = \frac{3}{2}$ and multiply to $\frac{c}{a} = \frac{-2}{2} = -1$. Test the pair.

$2 + \left(-\tfrac{1}{2}\right) = \tfrac{3}{2}, \qquad 2 \cdot \left(-\tfrac{1}{2}\right) = -1$

Both totals match, so the claim is correct.