Math / Algebra

QuestionSolving a word problem using a quadratic equation with rational roots
The length of a rectangle is 3 ft less than twice the width, and the area of the rectangle is $65 \mathrm{ft}^{2}$ . Find the dimensions of the rectangle.
Length : $\square$ ft
Width : $\square$ ft

Studdy Solution

STEP 1

1. The problem involves a rectangle with a specific relationship between its length and width.
2. The area of the rectangle is given as $65 \, \text{ft}^2$ .
3. The length is expressed in terms of the width.

STEP 2

1. Define variables for the dimensions of the rectangle.
2. Set up an equation for the area using the given relationship.
3. Solve the quadratic equation for the width.
4. Calculate the length using the width.
5. Verify the dimensions satisfy the area condition.

STEP 3

Define the variables: Let $w$ be the width of the rectangle in feet. The length $l$ is given as 3 feet less than twice the width, so:
$l = 2w - 3$

STEP 4

Set up the equation for the area of the rectangle. The area $A$ is given by:
$A = l \times w$
Substitute the expression for the length:
$65 = (2w - 3) \times w$

STEP 5

Expand and simplify the equation to form a quadratic equation:
$65 = 2w^2 - 3w$
Rearrange the equation to set it to zero:
$2w^2 - 3w - 65 = 0$

STEP 6

Solve the quadratic equation using the quadratic formula, where $a = 2$ , $b = -3$ , and $c = -65$ :
The quadratic formula is:
$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Substitute the values:
$w = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-65)}}{2 \cdot 2}$
$w = \frac{3 \pm \sqrt{9 + 520}}{4}$
$w = \frac{3 \pm \sqrt{529}}{4}$
$w = \frac{3 \pm 23}{4}$
Calculate the roots:
$w = \frac{3 + 23}{4} = \frac{26}{4} = 6.5$
$w = \frac{3 - 23}{4} = \frac{-20}{4} = -5$
Since width cannot be negative, we choose $w = 6.5$ .

STEP 7

Calculate the length using the width:
$l = 2w - 3$ $l = 2(6.5) - 3$ $l = 13 - 3$ $l = 10$

STEP 8

Verify the dimensions satisfy the area condition:
Area = $l \times w = 10 \times 6.5 = 65 \, \text{ft}^2$
The dimensions are verified.
The dimensions of the rectangle are:
Length: $10 \, \text{ft}$
Width: $6.5 \, \text{ft}$

Was this helpful?

QuestionSolving a word problem using a quadratic equation with rational roots
The length of a rectangle is 3 ft less than twice the width, and the area of the rectangle is $65 \mathrm{ft}^{2}$ . Find the dimensions of the rectangle.
Length : $\square$ ft
Width : $\square$ ft

Studdy Solution

STEP 1

1. The problem involves a rectangle with a specific relationship between its length and width.
2. The area of the rectangle is given as $65 \, \text{ft}^2$ .
3. The length is expressed in terms of the width.

STEP 2

1. Define variables for the dimensions of the rectangle.
2. Set up an equation for the area using the given relationship.
3. Solve the quadratic equation for the width.
4. Calculate the length using the width.
5. Verify the dimensions satisfy the area condition.

STEP 3

Define the variables: Let $w$ be the width of the rectangle in feet. The length $l$ is given as 3 feet less than twice the width, so:
$l = 2w - 3$

STEP 4

Set up the equation for the area of the rectangle. The area $A$ is given by:
$A = l \times w$
Substitute the expression for the length:
$65 = (2w - 3) \times w$

STEP 5

Expand and simplify the equation to form a quadratic equation:
$65 = 2w^2 - 3w$
Rearrange the equation to set it to zero:
$2w^2 - 3w - 65 = 0$

STEP 6

STEP 7

Calculate the length using the width:
$l = 2w - 3$ $l = 2(6.5) - 3$ $l = 13 - 3$ $l = 10$

STEP 8

Verify the dimensions satisfy the area condition:
Area = $l \times w = 10 \times 6.5 = 65 \, \text{ft}^2$
The dimensions are verified.
The dimensions of the rectangle are:
Length: $10 \, \text{ft}$
Width: $6.5 \, \text{ft}$

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionSolving a word problem using a quadratic equation with rational rootsThe length of a rectangle is 3 ft less than twice the width, and the area of the rectangle is 65ft265 \mathrm{ft}^{2}65ft2. Find the dimensions of the rectangle.Length : □\square□ ftWidth : □\square□ ft

Studdy Solution

STEP 1

1. The problem involves a rectangle with a specific relationship between its length and width.2. The area of the rectangle is given as 65 ft2 65 \, \text{ft}^2 65ft2.3. The length is expressed in terms of the width.

STEP 2

1. Define variables for the dimensions of the rectangle.2. Set up an equation for the area using the given relationship.3. Solve the quadratic equation for the width.4. Calculate the length using the width.5. Verify the dimensions satisfy the area condition.

STEP 3

Define the variables: Let w w w be the width of the rectangle in feet. The length l l l is given as 3 feet less than twice the width, so:l=2w−3 l = 2w - 3 l=2w−3

STEP 4

Set up the equation for the area of the rectangle. The area A A A is given by:A=l×w A = l \times w A=l×wSubstitute the expression for the length:65=(2w−3)×w 65 = (2w - 3) \times w 65=(2w−3)×w

STEP 5

Expand and simplify the equation to form a quadratic equation:65=2w2−3w 65 = 2w^2 - 3w 65=2w2−3wRearrange the equation to set it to zero:2w2−3w−65=0 2w^2 - 3w - 65 = 0 2w2−3w−65=0

STEP 6

STEP 7

Calculate the length using the width:l=2w−3 l = 2w - 3 l=2w−3 l=2(6.5)−3 l = 2(6.5) - 3 l=2(6.5)−3 l=13−3 l = 13 - 3 l=13−3 l=10 l = 10 l=10

STEP 8

Verify the dimensions satisfy the area condition:Area = l×w=10×6.5=65 ft2 l \times w = 10 \times 6.5 = 65 \, \text{ft}^2 l×w=10×6.5=65ft2The dimensions are verified. The dimensions of the rectangle are:Length: 10 ft 10 \, \text{ft} 10ftWidth: 6.5 ft 6.5 \, \text{ft} 6.5ft

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionSolving a word problem using a quadratic equation with rational roots
The length of a rectangle is 3 ft less than twice the width, and the area of the rectangle is $65 \mathrm{ft}^{2}$ . Find the dimensions of the rectangle.
Length : $\square$ ft
Width : $\square$ ft

1. The problem involves a rectangle with a specific relationship between its length and width.
2. The area of the rectangle is given as $65 \, \text{ft}^2$ .
3. The length is expressed in terms of the width.

1. Define variables for the dimensions of the rectangle.
2. Set up an equation for the area using the given relationship.
3. Solve the quadratic equation for the width.
4. Calculate the length using the width.
5. Verify the dimensions satisfy the area condition.

Define the variables: Let $w$ be the width of the rectangle in feet. The length $l$ is given as 3 feet less than twice the width, so:
$l = 2w - 3$

Set up the equation for the area of the rectangle. The area $A$ is given by:
$A = l \times w$
Substitute the expression for the length:
$65 = (2w - 3) \times w$

Expand and simplify the equation to form a quadratic equation:
$65 = 2w^2 - 3w$
Rearrange the equation to set it to zero:
$2w^2 - 3w - 65 = 0$

Calculate the length using the width:
$l = 2w - 3$ $l = 2(6.5) - 3$ $l = 13 - 3$ $l = 10$

Verify the dimensions satisfy the area condition:
Area = $l \times w = 10 \times 6.5 = 65 \, \text{ft}^2$
The dimensions are verified.
The dimensions of the rectangle are:
Length: $10 \, \text{ft}$
Width: $6.5 \, \text{ft}$