Math Problem Statement

A rectangular phone has a screen with a length that is 3 inches more than its width. The area of the screen is 50 square inches. Find the dimensions of the screen by solving the quadratic equation.

Solution

To find the dimensions of the rectangular phone screen, we'll use the information given to set up a quadratic equation.

Step-by-Step Solution

  1. Define Variables:

    • Let ww be the width of the screen in inches.
    • Since the length is 3 inches more than the width, the length ll can be expressed as l=w+3l = w + 3.
  2. Write the Area Equation:

    • The area of a rectangle is given by Area=length×width\text{Area} = \text{length} \times \text{width}.
    • Substituting the expressions for length and width, we get: w(w+3)=50w(w + 3) = 50
  3. Set Up the Quadratic Equation:

    • Expanding the equation, we have: w2+3w=50w^2 + 3w = 50
    • To form a standard quadratic equation, move all terms to one side: w2+3w50=0w^2 + 3w - 50 = 0
  4. Solve the Quadratic Equation:

    • To solve this quadratic equation, we can use the quadratic formula: w=b±b24ac2aw = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=1a = 1, b=3b = 3, and c=50c = -50.
  5. Calculate the Discriminant:

    • The discriminant (b24acb^2 - 4ac) is: 324(1)(50)=9+200=2093^2 - 4(1)(-50) = 9 + 200 = 209
  6. Apply the Quadratic Formula: w=3±2092w = \frac{-3 \pm \sqrt{209}}{2}

  7. Determine the Width:

    • Since width cannot be negative, we take the positive root: w=3+2092w = \frac{-3 + \sqrt{209}}{2}
    • Calculating the value: w3+14.456211.45625.728w \approx \frac{-3 + 14.456}{2} \approx \frac{11.456}{2} \approx 5.728
  8. Determine the Length:

    • The length ll is: l=w+35.728+3=8.728l = w + 3 \approx 5.728 + 3 = 8.728

Conclusion

The dimensions of the phone screen are approximately:

  • Width: 5.728 inches
  • Length: 8.728 inches

Would you like more details or have any questions?

Related Questions:

  1. How would the dimensions change if the area of the screen was 100 square inches instead?
  2. Can the quadratic formula be used for all quadratic equations? Why or why not?
  3. What are other methods to solve quadratic equations apart from the quadratic formula?
  4. How can you verify that the dimensions found are correct?
  5. What would happen if the length was 4 inches more than the width instead of 3?

Tip: Always check the discriminant of a quadratic equation before applying the quadratic formula to ensure real solutions exist.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Rectangles
Area Calculation

Formulas

Quadratic equation formula

Theorems

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Suitable Grade Level

Grades 7-9