The positive difference between two consecutive odd perfect squares is 336. Compute the larger of the two squares.
Let the first perfect square be x^2
then the next odd consecutive square = (x+2)^2
(x+2)^2 - x^2 = 336
take over
once you have x you will need (x+2)^2
Let's assume that the two consecutive odd perfect squares are n² and (n+2)².
The positive difference between the two squares is given as:
(n+2)² - n² = 336
Expanding the equation, we get:
(n² + 4n + 4) - n² = 336
Simplifying and combining like terms, we have:
4n + 4 = 336
Subtracting 4 from both sides:
4n = 332
Dividing by 4:
n = 83
The larger of the two squares is (n+2)² = (83+2)² = 85²
Therefore, the larger square is 85², which is equal to 7225.
To find the larger square, let's set up an equation based on the given information.
Let's assume the first odd perfect square is x^2. Since the squares are consecutive, the next odd perfect square would be (x + 2)^2.
We are given that the positive difference between these two squares is 336. So, we can write the equation as:
(x + 2)^2 - x^2 = 336
Expanding the left side of the equation:
x^2 + 4x + 4 - x^2 = 336
Simplifying the equation:
4x + 4 = 336
Subtracting 4 from both sides:
4x = 332
Dividing by 4:
x = 83
Now we have the value of x, which represents the first odd perfect square. To find the larger square, we substitute x into our assumption:
(x + 2)^2 = (83 + 2)^2 = 85^2
Therefore, the larger of the two squares is 85^2, which is equal to 7,225.