find three consecutive odd integers such that the sum of the first and the second is 201 less the square of the third
the numbers are x,x+2,x+4
x + x+2 = (x+4)^2 - 201
Now just expand, collect terms, and solve the quadratic.
Let's assume the three consecutive odd integers are represented by x, (x + 2), and (x + 4).
According to the problem, the sum of the first and second is equal to 201 less than the square of the third. Mathematically, we can write this as:
x + (x + 2) = (x + 4)^2 - 201
Now, let's solve this equation step by step:
1. Expand and simplify the right side of the equation:
x + x + 2 = (x^2 + 8x + 16) - 201
2x + 2 = x^2 + 8x + 16 - 201
2x + 2 = x^2 + 8x - 185
2. Move all terms to one side to obtain a quadratic equation:
0 = x^2 + 8x - 185 - 2x - 2
0 = x^2 + 6x - 187
3. Factor the quadratic equation:
0 = (x - 11)(x + 17)
Now, we have two possible values for x: x = 11 or x = -17.
4. Substitute each value of x to find the three consecutive odd integers:
For x = 11:
The three consecutive odd integers are 11, 13, and 15.
For x = -17:
The three consecutive odd integers are -17, -15, and -13.
Therefore, there are two sets of three consecutive odd integers that satisfy the given conditions: (11, 13, 15) and (-17, -15, -13).