find three consecutive positive odd integers such that the sum of the squares of the first and second integers is equal to the square of the third integer minus 7?
Let the numbers be x,x+2 and x+4.
x^2 + (x+2)^2 = (x+4)^2 - 7
x = 5
check:
25+49 = 81-7
yes
To find three consecutive positive odd integers that satisfy the given condition, let's make use of algebraic equations:
Let's assume the first odd integer as "x". The second odd integer would be "x + 2" since it is consecutive. And the third odd integer would be "x + 4".
Now, we can rewrite the given condition as an equation:
The sum of the squares of the first and second integers is equal to the square of the third integer minus 7.
x^2 + (x + 2)^2 = (x + 4)^2 - 7
Expanding this equation:
x^2 + x^2 + 4x + 4 = x^2 + 8x + 16 - 7
Combining like terms and simplifying the equation:
2x^2 + 4x + 4 = x^2 + 8x + 9
Rearranging the equation to set it equal to zero:
2x^2 + 4x + 4 - x^2 - 8x - 9 = 0
Simplifying further:
x^2 - 4x - 5 = 0
Now, we can solve this quadratic equation by factoring or using the quadratic formula. Let's use factoring to find the values of x.
Factoring the quadratic equation:
(x - 5)(x + 1) = 0
Setting each factor equal to zero:
x - 5 = 0 OR x + 1 = 0
Solving for x:
x = 5 OR x = -1
Since we are looking for positive integers, we can disregard the value of x = -1.
Therefore, the first odd integer is x = 5. The second odd integer is x + 2 = 5 + 2 = 7. And the third odd integer is x + 4 = 5 + 4 = 9.
Hence, the three consecutive positive odd integers that satisfy the given condition are 5, 7, and 9.