Question

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

Answer 1

To find the two integers, we set one as x and the other as y = 2x - 6. We formed a quadratic equation from the condition that the sum of their squares is 680 and solved it to find the values of x and y.

Step-by-step explanation:

Let's denote the smaller integer as x and the larger integer as y. According to the problem, one positive integer is 6 less than twice another, which translates to the equation y = 2x - 6. The sum of their squares is given as 680, so we write that as x^2 + y^2 = 680.

Substituting the expression for y from the first equation into the second, we get x^2 + (2x - 6)^2 = 680. Expanding this, we arrive at an equation in one variable: x^2 + 4x^2 - 24x + 36 = 680, which simplifies to 5x^2 - 24x - 644 = 0. This is a quadratic equation that can be factored or solved using the quadratic formula.

Factoring, we find the roots of the quadratic equation which will give us the value for x, and subsequently we can find y using the first equation. By solving, we determine that the integers that satisfy the given conditions are x and y.

Answer 2

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√(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`