Math Problem Statement

A rectangular garden has a length of 8 meters and a width of 5 meters. If you increase the length and width by the same amount, the area of the garden will double. What is the new length of the garden?

Solution

Let's define the increase in both the length and the width as xx meters.

  • The original length of the garden is 8 meters, and the original width is 5 meters.
  • The original area of the garden is 8×5=40m28 \times 5 = 40 \, \text{m}^2.
  • After increasing both dimensions by xx, the new length will be 8+x8 + x meters, and the new width will be 5+x5 + x meters.
  • The new area of the garden is (8+x)(5+x)(8 + x)(5 + x).
  • According to the problem, the new area is double the original area, so the new area is 2×40=80m22 \times 40 = 80 \, \text{m}^2.

Now, we set up the equation for the new area:

(8+x)(5+x)=80(8 + x)(5 + x) = 80

Expanding the left-hand side:

(8+x)(5+x)=40+8x+5x+x2=x2+13x+40(8 + x)(5 + x) = 40 + 8x + 5x + x^2 = x^2 + 13x + 40

Now, set this equal to 80:

x2+13x+40=80x^2 + 13x + 40 = 80

Subtract 80 from both sides:

x2+13x40=0x^2 + 13x - 40 = 0

We now solve this quadratic equation using the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For the equation x2+13x40=0x^2 + 13x - 40 = 0, we have:

  • a=1a = 1
  • b=13b = 13
  • c=40c = -40

Substituting into the quadratic formula:

x=13±1324(1)(40)2(1)x = \frac{-13 \pm \sqrt{13^2 - 4(1)(-40)}}{2(1)} x=13±169+1602x = \frac{-13 \pm \sqrt{169 + 160}}{2} x=13±3292x = \frac{-13 \pm \sqrt{329}}{2}

Since 32918.14\sqrt{329} \approx 18.14, we get:

x=13±18.142x = \frac{-13 \pm 18.14}{2}

This gives two solutions:

x=13+18.142=5.142=2.57x = \frac{-13 + 18.14}{2} = \frac{5.14}{2} = 2.57 x=1318.142=31.142=15.57x = \frac{-13 - 18.14}{2} = \frac{-31.14}{2} = -15.57

Since xx must be a positive value (as it represents a physical length increase), we take x=2.57x = 2.57.

Thus, the new length of the garden is:

8+x=8+2.57=10.57meters8 + x = 8 + 2.57 = 10.57 \, \text{meters}

The new length of the garden is approximately 10.57 meters.

Would you like more details on any step?

Here are five related questions you might find interesting:

  1. What would happen if the width of the garden were not increased?
  2. How does the quadratic formula help in real-world problems?
  3. What are other ways to solve quadratic equations besides the quadratic formula?
  4. How would the new width of the garden change if the area were tripled instead of doubled?
  5. What if only the length of the garden were increased to double the area?

Tip: Always check for the physical meaning of your solutions, especially when dealing with geometry, to ensure you choose realistic values.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Geometry
Area

Formulas

Area of rectangle: length × width
Quadratic equation formula: ax^2 + bx + c = 0

Theorems

Quadratic Formula

Suitable Grade Level

Grades 9-10