Question
original side length of a square when each side is lengthened by a certain number of inches and the new area is given
Answers
GPT-4o mini
To find the original side length of a square when each side is lengthened by a certain number of inches and the new area is given, we can use algebra.
Let's denote:
- \( s \): the original side length of the square
- \( x \): the number of inches each side is lengthened
- \( A \): the new area of the square after lengthening the sides
After lengthening each side by \( x \) inches, the new side length will be \( s + x \). The area of the square with this new side length can be expressed as:
\[
A = (s + x)^2
\]
Expanding this equation gives:
\[
A = s^2 + 2sx + x^2
\]
Now, we can relate this to the original area of the square, which is \( s^2 \). Rearranging the equation, we have:
\[
s^2 + 2sx + x^2 = A
\]
To find the original side length \( s \), we can rearrange this equation:
\[
s^2 + 2sx + (x^2 - A) = 0
\]
This is a quadratic equation in the standard form \( as^2 + bs + c = 0 \), where:
- \( a = 1 \)
- \( b = 2x \)
- \( c = x^2 - A \)
We can solve for \( s \) using the quadratic formula:
\[
s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Substituting \( a \), \( b \), and \( c \):
\[
s = \frac{-2x \pm \sqrt{(2x)^2 - 4 \cdot 1 \cdot (x^2 - A)}}{2 \cdot 1}
\]
Simplifying this, we have:
\[
s = \frac{-2x \pm \sqrt{4x^2 - 4(x^2 - A)}}{2}
\]
\[
s = \frac{-2x \pm \sqrt{4A}}{2}
\]
\[
s = -x \pm \sqrt{A}
\]
Since side lengths cannot be negative, we take the positive solution:
\[
s = \sqrt{A} - x
\]
This formula gives us the original side length of the square given the lengthening amount \( x \) and the new area \( A \).
Let's denote:
- \( s \): the original side length of the square
- \( x \): the number of inches each side is lengthened
- \( A \): the new area of the square after lengthening the sides
After lengthening each side by \( x \) inches, the new side length will be \( s + x \). The area of the square with this new side length can be expressed as:
\[
A = (s + x)^2
\]
Expanding this equation gives:
\[
A = s^2 + 2sx + x^2
\]
Now, we can relate this to the original area of the square, which is \( s^2 \). Rearranging the equation, we have:
\[
s^2 + 2sx + x^2 = A
\]
To find the original side length \( s \), we can rearrange this equation:
\[
s^2 + 2sx + (x^2 - A) = 0
\]
This is a quadratic equation in the standard form \( as^2 + bs + c = 0 \), where:
- \( a = 1 \)
- \( b = 2x \)
- \( c = x^2 - A \)
We can solve for \( s \) using the quadratic formula:
\[
s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Substituting \( a \), \( b \), and \( c \):
\[
s = \frac{-2x \pm \sqrt{(2x)^2 - 4 \cdot 1 \cdot (x^2 - A)}}{2 \cdot 1}
\]
Simplifying this, we have:
\[
s = \frac{-2x \pm \sqrt{4x^2 - 4(x^2 - A)}}{2}
\]
\[
s = \frac{-2x \pm \sqrt{4A}}{2}
\]
\[
s = -x \pm \sqrt{A}
\]
Since side lengths cannot be negative, we take the positive solution:
\[
s = \sqrt{A} - x
\]
This formula gives us the original side length of the square given the lengthening amount \( x \) and the new area \( A \).