Find 2 consecutive positive intergers such of the first and the square of the second is 19
To solve this problem, let's represent the two consecutive positive integers as "x" and "x + 1" where "x" is the smaller integer.
According to the problem, the first integer (x) and the square of the second integer ((x + 1)^2) sum up to 19.
So we can set up an equation based on this information:
x + (x + 1)^2 = 19
Let's solve this equation step by step:
Expand the square term:
x + (x^2 + 2x + 1) = 19
Combine like terms:
x + x^2 + 2x + 1 = 19
Rearrange the equation:
x^2 + 3x + 1 = 19
Subtract 19 from both sides:
x^2 + 3x - 18 = 0
Now, we can factor the quadratic equation if possible:
(x - 3)(x + 6) = 0
Setting each factor equal to zero, we get:
x - 3 = 0 or x + 6 = 0
Solving for x in each case:
x = 3 or x = -6
Since we are looking for positive consecutive integers, we can select x = 3.
Therefore, the two consecutive positive integers are 3 and 4.