Question

One positive integer is 6 less than twice another. The sum of their squares is 680 . Find the integers.

Answer 1

Let's call the two numbers x and y. The first sentence, one is 6 less than twice another, translates to

$x = 2y-6$

We also know the sum of their squares:

$x^2+y^2 = 680$

So, we have the following system:

$\begin{cases} x = 2y-6 \\ x^2+y^2 = 680 \end{cases}$

We can use the expression for x in terms of y from the first equation to turn the second equation into something involving y alone:

$x^2+y^2 = 680 \to (2y-6)^2+y^2 = 680$

Expand the square:

$5y^2-24y+36 = 680$

Subtract 680 from both sides:

$5y^2-24y-644 = 0$

Using the quadratic equation

$y_{1,2} = \cfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

we find the two solutions

$y_1 = -\cfrac{46}{5},\quad y_2 = 14$

Since we know that both numbers are integer we can only accept the second solution. It yields the following x value:

$x = 2x-6\ \land\ y = 14 \implies x = 2\cdot 14 - 6 = 28-6=22$