We are given a rectangular screen with a diagonal of 20 inches, and the width of the screen is 4 inches less than its length. To model this, we use the Pythagorean theorem, which states:

$\text{Diagonal}^2 = \text{Length}^2 + \text{Width}^2$

From the diagram:

• The diagonal is 20 inches.
• The length is $x$ inches.
• The width is $x - 4$ inches.

The equation modeling this situation is:

$x^2 + (x - 4)^2 = 20^2$

First, let's solve this equation step by step.

Step 1: Expand the equation

$x^2 + (x - 4)^2 = 400$

Expanding $(x - 4)^2$: $(x - 4)^2 = x^2 - 8x + 16$

So the equation becomes: $x^2 + x^2 - 8x + 16 = 400$

Step 2: Simplify the equation

Combine like terms: $2x^2 - 8x + 16 = 400$

Step 3: Move all terms to one side of the equation

$2x^2 - 8x + 16 - 400 = 0$

$2x^2 - 8x - 384 = 0$

Step 4: Divide the entire equation by 2 to simplify

$x^2 - 4x - 192 = 0$

Step 5: Solve using the quadratic formula

The quadratic formula is given by:

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

For the equation $x^2 - 4x - 192 = 0$, $a = 1$, $b = -4$, and $c = -192$.

Let's calculate this.The solutions to the quadratic equation are $x = 16$ and $x = -12$. Since the length of the screen cannot be negative, the valid solution is:

$x = 16$

Thus, the length of the screen is 16 inches.

Would you like more details or further clarification?

Here are 5 related questions for further practice:

1. How would you find the width of the screen using the value of $x$?
2. What would the area of the screen be, using the length and width?
3. How can the Pythagorean theorem be applied to other shapes?
4. What are some real-world applications of the quadratic formula?
5. How would the equation change if the width was a different amount less than the length?

Tip: When solving a quadratic equation, always check for extraneous or invalid solutions, especially in real-world problems where certain values (like negative lengths) don't make sense.