Find two consecutive positive integers such that the square of the first equals 25 more than three times the second.

Answer 1

We're given: y - x = 1, x > 0, y > 0

We can write the second part into an algebraic equation: x² = 3y + 25

If y - x = 1, then y = 1 + x

Now we plug (1 + x) in for y and we have: x² = 3(1 + x) + 25

Let's simplify: x² = 3 + 3x + 25 = 3x + 28

Now let's make this a quadratic equation and solve for x: x² - 3x - 28 = 0

Now using the quadratic formula, we get x = 7 or -4

Since x must be positive, we choose x = 7

Now we have y = 1 + 7 = 8

So our two integers are 7 & 8