One positive integer is 2 less than twice another. The sum of their squares is 628. Find the integers

Question

 looking for the two integers

Answer 1

The statement of the problem is easily converted to two equations, the first of which will be x = 2y -2.

Write the second equation based on the second sentence.

 

Now you have a system of two equations with two unknowns. Substitute into the second equation,u sing the first equation.

That gives you a quadratic equation in y. Solve that!

Answer 2

y=first number

x=2y-2 (second number)

y^2+(2y-2)^2=628

y^2+4y^2-8y+4=628

5y^2-8y+4-628=0

5y^2-8y-624=0

factor the equation

5*624=5*2*2*2*2*3*13

now group the factors so that the two products have a difference of 8, the coefficient of the middle term

5*2*2*2*2*3*13=(2*2*3*5)*(2*2*13)=60*52 and 60-52=8

(5y    )(y    )=0 so far

(5y    )(y   12)=0 so far

(5y  52)(y   12)=0 so far

put the signs in correctly

(5y+52)(y-12)=0

y-12=0  (forget about 5y+12 because it yields a negative answer)

y-12=0

y=12

x=2*12-2

x=24-2

x=22

the numbers are 12 and 22

22 is 2 less than 24 and 12^2+22^2=144+484=628