One positive integer is 5 less than twice another. The sum of their squares is 725. Find the integers.

Answer 1

We know that M and N=2M-5 are the two integers that we want to find and we also know that M^2+N^2=725. Substituting N=2M-5 into the preceding equation then gives us

M^2+(2M-5)^2=725

M^2+((2M)^2-2(2M)(5)+5^2)=725

M^2+(4M^2-20M+25)=725

5M^2-20M+25=725

(5M^2-20M+25)/5=725/5

M^2-4M+5=145

M^2-4M+5-145=145-145

M^2-4M-140=0

(M-14)(M+10)=0 (by standard factorization strategies)

⇒ M-14=0 or M+10=0

⇒ M-14+14=0+14 or M+10-10=0-10

⇒ M=14 or M=-10

⇒ M=14 since M is positive

⇒ N=2(14)-5=23

Thus the two integers are 14 and 23.

Answer 2

Hi.

Let's start by assigning the letter "x" to the first integer and the letter "y" to the second integer.

The problem tells us that the sum of the squares of these two integers is 725. So:

x2 + y2 = 725

We're also told that the first integer (x) is "5 less than twice" the second integer (y). So:

x = 2y - 5

Therefore:

x² + y² = 725

(2y-5)² + y² = 725

[(2y - 5) (2y - 5)] + y² = 725

4y² -20y + 25 + y² = 725

5y² - 20y - 700 = 0

Now we divide by 5

y² - 4y - 140 = 0

Next, we factor:

(y + 10) (y - 14) = 0

Solve separately for each:

y = -10, y = 14

The answer has to be a positive integer, so let's choose y = 14

If y = 14, then x = 2y -5 = 2(14) - 5 = 23

Let's see if we are right. Does 14² + 23² = 725?

14² = 195

23² = 529

195 + 529 = 724

We have a discrepancy of 1. But we need to stick to whole numbers so we are going to go with our answers.