find two consecutive integers such that the sum of their squares is 145

Question

i have a test tmr and cannot figure this problem out to save my life. ive looked at it and have tried to figure out #s that square and equal 145 but i honestly have no idea. HELP!!!

Answer 1

1 st integer = x

2nd integer = x+1

 

x²+(x+1)² =145

 

x² +x²+2x+1 =145

2x²+2x=144

2x²+2x-144=0

x²+x-72=0

(x+9)(x-8)=0

x=-9  and x=8

 

so

1 st integer = ±8
2nd integer = ±9

 

 

Answer 2

Lets call our integers x and y.

 

How do we represent the "sum of their squares" with a mathematical expression?

 

This would be x² + y² and we are told this is equal to 145.

 

But I can't solve one equation with two variables. x² + y² = 145 has many solutions. The trick is in how you represent "consecutive integers." They come one after the other such as (4,5), (26,27) etc. How do we get from one number to the next consecutive one? We add one. So if my first number is x, the next consecutive number is x + 1. So lets replace the y in our equation with (x+1)

 

x² + (x+1)² = 145

 

Now we have an equation with one variable that I can solve.

 

Foil first to get x² +x² +2x + 1 = 145

 

2x² + 2x +1 = 145

 

Since this is a quadratic equation we can use the quadratic formula or factor to solve it.

 

First get everything to one side 2x² + 2x - 144 = 0

 

and you will get that x = 8 or -9

 

These are the two possible answers for the first consecutive integer x. Then we can figure out the next consecutive integer (y) by adding one to each. So there are two sets of answers, 8 and 9, or -9 and -8. A simple check shows they both will give you 145.