Math Problem Statement

A garden;s length is 8 meters more than its width. If both the width and the length of the garden are doubled, the area is 192m^2. What are the dimensions of the length and width?

Solution

Let's denote the width of the garden as ww meters. Since the length is 8 meters more than the width, the length can be expressed as l=w+8l = w + 8.

Now, according to the problem, both the width and the length are doubled, and the area of the garden becomes 192 square meters. The area of a rectangle is given by the formula:

Area=Length×Width\text{Area} = \text{Length} \times \text{Width}

After doubling the dimensions, the width becomes 2w2w and the length becomes 2(w+8)2(w + 8). The new area is 192 square meters, so:

2w×2(w+8)=1922w \times 2(w + 8) = 192

Simplify the equation:

4w(w+8)=1924w(w + 8) = 192

Divide both sides by 4:

w(w+8)=48w(w + 8) = 48

Now, expand the equation:

w2+8w=48w^2 + 8w = 48

Rearrange it into a standard quadratic form:

w2+8w48=0w^2 + 8w - 48 = 0

Now, solve this quadratic equation using the quadratic formula. The quadratic formula is:

w=b±b24ac2aw = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For the equation w2+8w48=0w^2 + 8w - 48 = 0, the coefficients are:

  • a=1a = 1
  • b=8b = 8
  • c=48c = -48

Substitute these values into the quadratic formula:

w=8±824(1)(48)2(1)w = \frac{-8 \pm \sqrt{8^2 - 4(1)(-48)}}{2(1)} w=8±64+1922w = \frac{-8 \pm \sqrt{64 + 192}}{2} w=8±2562w = \frac{-8 \pm \sqrt{256}}{2} w=8±162w = \frac{-8 \pm 16}{2}

Now, solve for ww:

  1. w=8+162=82=4w = \frac{-8 + 16}{2} = \frac{8}{2} = 4
  2. w=8162=242=12w = \frac{-8 - 16}{2} = \frac{-24}{2} = -12

Since the width cannot be negative, w=4w = 4 meters.

Now, substitute w=4w = 4 into the expression for the length:

l=w+8=4+8=12metersl = w + 8 = 4 + 8 = 12 \, \text{meters}

Therefore, the original dimensions of the garden are:

  • Width = 4 meters
  • Length = 12 meters

Would you like further details or explanations?

Related Questions:

  1. What would the area of the garden be before doubling the dimensions?
  2. How would the problem change if the length was 10 meters more than the width?
  3. What if the area of the garden became 240 m² instead of 192 m² after doubling the dimensions?
  4. How do you solve a quadratic equation using the factorization method?
  5. Can you solve for the dimensions if the length is given as 2 meters more than twice the width?

Tip:

When solving quadratic equations, always check for common factors before applying the quadratic formula—it might simplify the process!

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Math Problem Analysis

Mathematical Concepts

Algebra
Geometry
Quadratic Equations

Formulas

Area of a rectangle: Area = Length × Width
Quadratic equation formula: ax^2 + bx + c = 0

Theorems

Quadratic Formula

Suitable Grade Level

Grades 8-10