To solve for the lengths of $JK$ and $KL$, we need to use the fact that point $K$ lies between points $J$ and $L$. This implies that:

$JK + KL = JL$

Given:

• $JK = x^2 - 4x$
• $KL = 3x - 2$
• $JL = 28$

We can set up the equation:

$(x^2 - 4x) + (3x - 2) = 28$

Simplifying:

$x^2 - 4x + 3x - 2 = 28$

$x^2 - x - 2 = 28$

Next, subtract 28 from both sides:

$x^2 - x - 30 = 0$

Now, solve this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For $x^2 - x - 30 = 0$, $a = 1$, $b = -1$, and $c = -30$. Substituting these values:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-30)}}{2(1)}$

$x = \frac{1 \pm \sqrt{1 + 120}}{2}$

$x = \frac{1 \pm \sqrt{121}}{2}$

$x = \frac{1 \pm 11}{2}$

So, $x$ can be:

$x = \frac{12}{2} = 6 \quad \text{or} \quad x = \frac{-10}{2} = -5$

Case 1: $x = 6$

Substitute $x = 6$ into the expressions for $JK$ and $KL$:

$JK = 6^2 - 4(6) = 36 - 24 = 12$

$KL = 3(6) - 2 = 18 - 2 = 16$

Check if $JK + KL = JL$:

$12 + 16 = 28$

This is correct.

Case 2: $x = -5$

Substitute $x = -5$:

$JK = (-5)^2 - 4(-5) = 25 + 20 = 45$

$KL = 3(-5) - 2 = -15 - 2 = -17$

However, a negative length for $KL$ is not physically meaningful, so $x = -5$ is not a valid solution.

Conclusion

The valid solution is $x = 6$, so:

$JK = 12, \quad KL = 16$

Do you want details or have any questions?

Here are 5 related questions:

1. How would the equation change if point K was outside the segment JL?
2. What happens if we had a cubic equation instead of a quadratic one?
3. How can we geometrically represent this situation?
4. What is the significance of discarding the negative root in this problem?
5. How do you solve quadratic equations by factoring?

Tip: Always check the physical meaning of your solutions, especially when dealing with geometric problems.