We are given a parallelogram $ABCD$, and diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E$. We know the following:

• $BE = x^2 - 21$
• $DE = 4x$

In any parallelogram, the diagonals bisect each other. This means that $BE = DE$.

Step-by-step Solution:

1. Since diagonals bisect each other, we can set up the equation: $BE = DE$ Substituting the expressions for $BE$ and $DE$: $x^2 - 21 = 4x$

2. Rearrange the equation to form a quadratic: $x^2 - 4x - 21 = 0$

3. Solve this quadratic equation using the quadratic formula: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-21)}}{2(1)}$ Simplifying the discriminant: $x = \frac{4 \pm \sqrt{16 + 84}}{2}$ $x = \frac{4 \pm \sqrt{100}}{2}$ $x = \frac{4 \pm 10}{2}$

4. This gives two possible solutions for $x$: $x = \frac{4 + 10}{2} = 7 \quad \text{or} \quad x = \frac{4 - 10}{2} = -3$

5. Since $x$ represents a length, it must be positive, so we take $x = 7$.

6. Now, substitute $x = 7$ back into the expression for $BD$. Since $BD = BE + DE$ and $BE = DE$, we can find: $BD = BE + DE = 2 \times BE = 2 \times (x^2 - 21)$ Substituting $x = 7$: $BD = 2 \times (7^2 - 21) = 2 \times (49 - 21) = 2 \times 28 = 56$

Thus, the length of $BD$ is $56$ units.

Would you like further details or have any questions? Here are some related questions you might find interesting:

1. What are the properties of a parallelogram's diagonals?
2. How do you solve a quadratic equation using factoring instead of the quadratic formula?
3. If the diagonals of a parallelogram bisect each other, what other geometric shapes share this property?
4. How does changing the value of $BE$ or $DE$ affect the solution for $BD$?
5. Can the quadratic formula ever result in two valid positive lengths for this kind of problem?

Tip: Always verify your solutions by plugging them back into the original equation to ensure they satisfy the problem's constraints.