Math / Algebra

QuestionQuestion Watch Video Show Examples
The width of a rectangle is 4 units less than the length. The area of the rectangle is 32 square units. What is the length, in units, of the rectangle?

Studdy Solution

STEP 1

What is this asking? We need to find the length of a rectangle whose width is 4 less than its length, and whose area is 32. Watch out! Don't mix up length and width!
Also, remember area is length times width, not length plus width!

STEP 2

1. Set up variables
2. Express the width
3. Write the area equation
4. Solve for the length

STEP 3

Let's call the length of the rectangle $L$ and the width $W$ .
Easy peasy!

STEP 4

The problem tells us the width is 4 less than the length.
We can write this mathematically as $W = L - 4$ .
Boom!

STEP 5

We know the area of a rectangle is length times width.
So, we can write the equation: $A = L \cdot W$ .

STEP 6

We're given that the area is 32.
And we just figured out that $W = L - 4$ .
Let's substitute these into our area equation: $32 = L \cdot (L - 4)$ .
Look at us go!

STEP 7

Now, let's distribute $L$ in the equation: $32 = L^2 - 4L$ .
This looks like a quadratic equation!

STEP 8

To solve a quadratic equation, we usually want it to equal zero.
So, let's subtract 32 from both sides: $0 = L^2 - 4L - 32$ .

STEP 9

Time to factor!
We're looking for two numbers that multiply to -32 and add up to -4.
Those magic numbers are -8 and 4.
So, we can rewrite our equation as $0 = (L - 8)(L + 4)$ .

STEP 10

Now, if $(L - 8)(L + 4) = 0$ , then either $L - 8 = 0$ or $L + 4 = 0$ .

STEP 11

If $L - 8 = 0$ , then by adding 8 to both sides we get $L = 8$ .
If $L + 4 = 0$ , then by adding -4 to both sides we get $L = -4$ .

STEP 12

Since length can't be negative (imagine a rectangle with a negative side!), we know the length must be $L = 8$ .
Woohoo!

STEP 13

The length of the rectangle is 8 units.

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QuestionQuestion Watch Video Show Examples
The width of a rectangle is 4 units less than the length. The area of the rectangle is 32 square units. What is the length, in units, of the rectangle?

Studdy Solution

STEP 1

What is this asking? We need to find the length of a rectangle whose width is 4 less than its length, and whose area is 32. Watch out! Don't mix up length and width!
Also, remember area is length times width, not length plus width!

STEP 2

1. Set up variables
2. Express the width
3. Write the area equation
4. Solve for the length

STEP 3

Let's call the length of the rectangle $L$ and the width $W$ .
Easy peasy!

STEP 4

The problem tells us the width is 4 less than the length.
We can write this mathematically as $W = L - 4$ .
Boom!

STEP 5

We know the area of a rectangle is length times width.
So, we can write the equation: $A = L \cdot W$ .

STEP 6

We're given that the area is 32.
And we just figured out that $W = L - 4$ .
Let's substitute these into our area equation: $32 = L \cdot (L - 4)$ .
Look at us go!

STEP 7

Now, let's distribute $L$ in the equation: $32 = L^2 - 4L$ .
This looks like a quadratic equation!

STEP 8

To solve a quadratic equation, we usually want it to equal zero.
So, let's subtract 32 from both sides: $0 = L^2 - 4L - 32$ .

STEP 9

Time to factor!
We're looking for two numbers that multiply to -32 and add up to -4.
Those magic numbers are -8 and 4.
So, we can rewrite our equation as $0 = (L - 8)(L + 4)$ .

STEP 10

Now, if $(L - 8)(L + 4) = 0$ , then either $L - 8 = 0$ or $L + 4 = 0$ .

STEP 11

If $L - 8 = 0$ , then by adding 8 to both sides we get $L = 8$ .
If $L + 4 = 0$ , then by adding -4 to both sides we get $L = -4$ .

STEP 12

Since length can't be negative (imagine a rectangle with a negative side!), we know the length must be $L = 8$ .
Woohoo!

STEP 13

The length of the rectangle is 8 units.

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QuestionQuestion Watch Video Show ExamplesThe width of a rectangle is 4 units less than the length. The area of the rectangle is 32 square units. What is the length, in units, of the rectangle?

Studdy Solution

STEP 1

What is this asking? We need to find the length of a rectangle whose width is 4 less than its length, and whose area is 32. Watch out! Don't mix up length and width!Also, remember area is length *times* width, not length *plus* width!

STEP 2

1. Set up variables2. Express the width3. Write the area equation4. Solve for the length

STEP 3

Let's call the length of the rectangle LLL and the width WWW.Easy peasy!

STEP 4

The problem tells us the width is **4 less than the length**.We can write this mathematically as W=L−4W = L - 4W=L−4.Boom!

STEP 5

We know the **area of a rectangle** is length times width.So, we can write the equation: A=L⋅WA = L \cdot WA=L⋅W.

STEP 6

We're given that the area is **32**.And we just figured out that W=L−4W = L - 4W=L−4.Let's substitute these into our area equation: 32=L⋅(L−4)32 = L \cdot (L - 4)32=L⋅(L−4).Look at us go!

STEP 7

Now, let's **distribute** LLL in the equation: 32=L2−4L32 = L^2 - 4L32=L2−4L.This looks like a **quadratic equation**!

STEP 8

To solve a quadratic equation, we usually want it to equal zero.So, let's **subtract 32** from both sides: 0=L2−4L−320 = L^2 - 4L - 320=L2−4L−32.

STEP 9

Time to **factor**!We're looking for two numbers that multiply to -32 and add up to -4.Those magic numbers are -8 and 4.So, we can rewrite our equation as 0=(L−8)(L+4)0 = (L - 8)(L + 4)0=(L−8)(L+4).

STEP 10

Now, if (L−8)(L+4)=0(L - 8)(L + 4) = 0(L−8)(L+4)=0, then either L−8=0L - 8 = 0L−8=0 or L+4=0L + 4 = 0L+4=0.

STEP 11

If L−8=0L - 8 = 0L−8=0, then by **adding 8** to both sides we get L=8L = 8L=8.If L+4=0L + 4 = 0L+4=0, then by **adding -4** to both sides we get L=−4L = -4L=−4.

STEP 12

Since length can't be negative (imagine a rectangle with a negative side!), we know the **length** must be L=8L = 8L=8.Woohoo!

STEP 13

The length of the rectangle is **8 units**.

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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.

QuestionQuestion Watch Video Show Examples
The width of a rectangle is 4 units less than the length. The area of the rectangle is 32 square units. What is the length, in units, of the rectangle?

What is this asking? We need to find the length of a rectangle whose width is 4 less than its length, and whose area is 32. Watch out! Don't mix up length and width!
Also, remember area is length times width, not length plus width!

1. Set up variables
2. Express the width
3. Write the area equation
4. Solve for the length

Let's call the length of the rectangle $L$ and the width $W$ .
Easy peasy!

The problem tells us the width is 4 less than the length.
We can write this mathematically as $W = L - 4$ .
Boom!

We know the area of a rectangle is length times width.
So, we can write the equation: $A = L \cdot W$ .

We're given that the area is 32.
And we just figured out that $W = L - 4$ .
Let's substitute these into our area equation: $32 = L \cdot (L - 4)$ .
Look at us go!

Now, let's distribute $L$ in the equation: $32 = L^2 - 4L$ .
This looks like a quadratic equation!

To solve a quadratic equation, we usually want it to equal zero.
So, let's subtract 32 from both sides: $0 = L^2 - 4L - 32$ .

Time to factor!
We're looking for two numbers that multiply to -32 and add up to -4.
Those magic numbers are -8 and 4.
So, we can rewrite our equation as $0 = (L - 8)(L + 4)$ .

Now, if $(L - 8)(L + 4) = 0$ , then either $L - 8 = 0$ or $L + 4 = 0$ .

If $L - 8 = 0$ , then by adding 8 to both sides we get $L = 8$ .
If $L + 4 = 0$ , then by adding -4 to both sides we get $L = -4$ .

Since length can't be negative (imagine a rectangle with a negative side!), we know the length must be $L = 8$ .
Woohoo!

The length of the rectangle is 8 units.