Please help me!!

Question

One positive integer is 4 less than twice another. The sum of their squares is 772 . Find the integers.

Answer 1

Let x and y be the integers. From the description:

y=2x-4

x²+y²=772

x²+4x²-16x+16 = 772

5x²-16x-756 = 0

by the quadratic equation

x = 14, -10.8 Use 14 since we are looking for a positive integer

y = 2(14)-4 = 24

x=14 and y = 24

Answer 2

14 and 24

24 = 2(14) -4 = 28-4 = 24

14^2 + 24^2 = 196 + 576 = 772

x = 2y -4

x^2 + y^2 = 772

(2y-4)^2 +y^2 = 772

4y^2 -16y + 16 + y^2 = 772

5y^2 -16y - 756 = 0

factor

(5y -54)(y -14) = 0

y-14 = 0

y = 14

x =2(14)-4 = 24

numbers are 14 and 24

Answer 3

We will call this positive integer x.

We will call the other integer y.

x=2y-4

x²+y²=772

We are looking for only integer solutions to the following equations.

First, plug in the first equation into the second,

(2y-4)²+y²=772

4y²-16y+16+y²=772

5y²-16y+16=772

5y²-16y-756=0

Use the quadratic formula:

y = 16±√(256-4(5)(-756))/(10)

y=(16±124)/(10)

We only want positive integers,

y=(16+124)/10

y=14

Plug back into the first equation,

x=2(14)-4

x=28-4

x=24.

Therefore, the two positive integer solutions to this system of equations are x=24 and y=14.