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Answer :
To find the length and width of the rectangular bedroom, let's use the information given:
1. Understanding the problem:
- The length of the bedroom is 2 feet more than its width.
- The area of the bedroom is 120 square feet.
2. Let’s set up the problem mathematically:
- Let the width be [tex]\( w \)[/tex] feet.
- Then the length will be [tex]\( w + 2 \)[/tex] feet (since it's 2 feet more than the width).
3. Write the equation for the area:
- The area of a rectangle is calculated as length times width.
- Thus, the equation becomes:
[tex]\[
(w + 2) \times w = 120
\][/tex]
4. Expand and rearrange the equation:
- Expand the left side of the equation:
[tex]\[
w^2 + 2w = 120
\][/tex]
- Subtract 120 from both sides to set the equation to zero:
[tex]\[
w^2 + 2w - 120 = 0
\][/tex]
5. Solve the quadratic equation:
- The equation [tex]\( w^2 + 2w - 120 = 0 \)[/tex] is in the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -120 \)[/tex].
- Use the quadratic formula to find [tex]\( w \)[/tex]:
[tex]\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
- Compute the discriminant:
[tex]\[
b^2 - 4ac = 2^2 - 4 \times 1 \times (-120) = 4 + 480 = 484
\][/tex]
- Find the square root of the discriminant, which is [tex]\(\sqrt{484} = 22\)[/tex].
- Plug the values into the quadratic formula:
[tex]\[
w = \frac{-2 \pm 22}{2}
\][/tex]
6. Calculate the possible values for [tex]\( w \)[/tex]:
- [tex]\( w = \frac{-2 + 22}{2} = \frac{20}{2} = 10 \)[/tex]
- [tex]\( w = \frac{-2 - 22}{2} = \frac{-24}{2} = -12 \)[/tex] (not a valid width since width cannot be negative)
7. Determine the width and length:
- So, the width is [tex]\( 10 \)[/tex] feet.
- The length, being 2 feet more than the width, is:
[tex]\[
w + 2 = 10 + 2 = 12 \text{ feet}
\][/tex]
Therefore, the bedroom's width is 10 feet and the length is 12 feet.
1. Understanding the problem:
- The length of the bedroom is 2 feet more than its width.
- The area of the bedroom is 120 square feet.
2. Let’s set up the problem mathematically:
- Let the width be [tex]\( w \)[/tex] feet.
- Then the length will be [tex]\( w + 2 \)[/tex] feet (since it's 2 feet more than the width).
3. Write the equation for the area:
- The area of a rectangle is calculated as length times width.
- Thus, the equation becomes:
[tex]\[
(w + 2) \times w = 120
\][/tex]
4. Expand and rearrange the equation:
- Expand the left side of the equation:
[tex]\[
w^2 + 2w = 120
\][/tex]
- Subtract 120 from both sides to set the equation to zero:
[tex]\[
w^2 + 2w - 120 = 0
\][/tex]
5. Solve the quadratic equation:
- The equation [tex]\( w^2 + 2w - 120 = 0 \)[/tex] is in the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -120 \)[/tex].
- Use the quadratic formula to find [tex]\( w \)[/tex]:
[tex]\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
- Compute the discriminant:
[tex]\[
b^2 - 4ac = 2^2 - 4 \times 1 \times (-120) = 4 + 480 = 484
\][/tex]
- Find the square root of the discriminant, which is [tex]\(\sqrt{484} = 22\)[/tex].
- Plug the values into the quadratic formula:
[tex]\[
w = \frac{-2 \pm 22}{2}
\][/tex]
6. Calculate the possible values for [tex]\( w \)[/tex]:
- [tex]\( w = \frac{-2 + 22}{2} = \frac{20}{2} = 10 \)[/tex]
- [tex]\( w = \frac{-2 - 22}{2} = \frac{-24}{2} = -12 \)[/tex] (not a valid width since width cannot be negative)
7. Determine the width and length:
- So, the width is [tex]\( 10 \)[/tex] feet.
- The length, being 2 feet more than the width, is:
[tex]\[
w + 2 = 10 + 2 = 12 \text{ feet}
\][/tex]
Therefore, the bedroom's width is 10 feet and the length is 12 feet.
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