SOLUTION: find two consecutive whole numbers such that the sum of their square is 41.

Question 515261: find two consecutive whole numbers such that the sum of their square is 41.
Answer by Maths68(1474) (Show Source): You can put this solution on YOUR website!
Let
One number = x
Other number = x+1
sum of their square = 41
x^2+(x+1)^=41
x^2+x^2+2x+1=41
2x^2+2x+1-41=0
2x^2+2x-40=0
2*(x^2+x-20)=0
2*(x^2+x-20)/2=0/2
x^2+x-20=0
x^2-4x+5x-20=0
x(x-4)+5(x-4)=0
(x-4)(x+5)=0
x-4=0 or x+5=o
x=4 or x=-5
One number = x = 4
Other number = x+1 = 4+1 = 5
4^2+5^2=41
16+25=41
41=41

RELATED QUESTIONS
FIND TWO CONSECUTIVE WHOLE NUMBERS SUCH THAT THE SUM OF THEIR SQUARES IS... (answered by Abbey,longjonsilver)

Find two consecutive whole numbers such that the sum of their squares is... (answered by flame8855)

Find two consecutive positive integers such that the sum of their squares is... (answered by algebrahouse.com)

What are two consecutive odd whole numbers such that the sum of their squares is... (answered by ankor@dixie-net.com)

Find three consecutive odd whole numbers such that the sum of the squares of the two... (answered by robertb)

find two consecutive positive even numbers such that the sum of their squares is 164. (answered by Alan3354)

find two consecutive positive numbers such that the sum of their squares is equal to... (answered by Big Poop)

Find two consecutive even natural numbers such that the sum of their squares is 340 . (answered by Theo)

Find two consecutive positive odd numbers such that the sum of their squares is... (answered by rothauserc)