You're on a roll — Step 1: Substitute the given root into the equation to find $k$. If $x = -2$ is a root of $x^2 + kx + 8 = 0$, then substituting $x=-2$ must satisfy the equation: $$(-2)^2 + k(-2) + 8 = 0$$ $$4 - 2k + 8 = 0$$ $$12 - 2k = 0$$ $$2k = 12$$ $$k = 6$$ Step 2: Write the complete quadratic equation. Substitute $k=6$ back into the original equation: $$x^2 + 6x + 8 = 0$$ Step 3: Find the other root using the sum of roots property. For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-\frac{b}{a}$. In our equation, $x^2 + 6x + 8 = 0$, we have $a=1$, $b=6$, $c=8$. Let the two roots be $x_1$ and $x_2$. We are given $x_1 = -2$. The sum of the roots is $x_1 + x_2 = -\frac{6}{1} = -6$. Substitute $x_1 = -2$: $$-2 + x_2 = -6$$ $$x_2 = -6 + 2$$ $$x_2 = -4$$ Alternatively, using the product of roots property: The product of the roots is $x_1 \cdot x_2 = \frac{c}{a} = \frac{8}{1} = 8$. Substitute $x_1 = -2$: $$-2 \cdot x_2 = 8$$ $$x_2 = \frac{8}{-2}$$ $$x_2 = -4$$ The other root is $\boxed{-4}$. Step 1: Define the two numbers. Let the first number be $x$. The second number is greater than the first by 4, so the second number is $x+4$. Step 2: Formulate the quadratic equation based on their product. The product of the two numbers is 21. $$x(x+4) = 21$$ Step 3: Expand and rearrange the equation into standard quadratic form. $$x^2 + 4x = 21$$ $$x^2 + 4x - 21 = 0$$ The quadratic equation is $\boxed{x^2 + 4x - 21 = 0}$. Step 4: Solve the quadratic equation by factoring. We need two numbers that multiply to -21 and add to 4. These numbers are 7 and -3. $$(x+7)(x-3) = 0$$ Step 5: Find the possible values for $x$. Set each factor to zero: $$x+7 = 0 \implies x = -7$$ $$x-3 = 0 \implies x = 3$$ Step 6: Determine the two numbers for each possible value of $x$. Case 1: If $x = 3$. The first number is 3. The second number is $x+4 = 3+4 = 7$. Check: $3 \times 7 = 21$. Case 2: If $x = -7$. The first number is -7. The second number is $x+4 = -7+4 = -3$. Check: $(-7) \times (-3) = 21$. The two numbers are either $\boxed{\text{3 and 7}}$ or $\boxed{\text{-7 and -3}}$. Got more? Send 'em.