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After on side of a square was lengthened by 2 inches and the other side was lengthened by 3 inches the area of the shape was doubled. What was the length of the original square

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Answer: 6 inches

Explanation:

Let's assume the original side length of the square is "x" inches.

After one side was lengthened by 2 inches, the new length of that side becomes "x + 2" inches.

After the other side was lengthened by 3 inches, the new length of that side becomes "x + 3" inches.

The area of the original square is given by: Area = x * x = x^2 square inches.

The area of the new shape (after lengthening the sides) is given by: Area = (x + 2) * (x + 3) = (x^2 + 5x + 6) square inches.

According to the problem, the area of the new shape is double the area of the original square. So, we can write the equation:

2 * (x^2) = x^2 + 5x + 6

Now, let's solve for "x" by simplifying the equation:

2x^2 = x^2 + 5x + 6

Subtract x^2 and 5x from both sides:

2x^2 - x^2 - 5x = 6

Simplify:

x^2 - 5x - 6 = 0

Now, we have a quadratic equation in standard form. To solve for "x," we can factor or use the quadratic formula. Factoring gives:

(x - 6)(x + 1) = 0

Setting each factor to zero and solving for "x":

x - 6 = 0 or x + 1 = 0

If x - 6 = 0, then x = 6.

If x + 1 = 0, then x = -1.

Since side lengths cannot be negative, the length of the original square is "x = 6 inches."

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