Question

Find three consecutive positive integers such that the sum of the squares of the first and second equals the square of the third.

Answer 1

please help?

Answer 2

let x = first integer
x + 1 = second integer
x + 2 = third integer

sum of the squares of the first and second equals the square of the third so

\[(x)^2 + (x+1)^2 = (x+2)^2\]
\[\implies x^2 + x^2 + 2x + 1 = x^2 + 4x + 4\]
\[ \implies 2x ^ 2 + 2x + 1 = x^2 + 4x + 4\]
\[\implies 2x ^2 - x^2 + 2x - 4x + 1 - 4 = 0\]
\[\implies x^2 - 2x - 3= 0\]
\[\implies (x-3)(x+1) = 0\]
therefore
x = 3 and x = -1
does that help?

Answer 3

yes thank you but, how did you get 2x from? and how are you combining x^2 and x^2 and getting 2x^2? can you explain please

Answer 4

which 2x?

Answer 5

in the second line of the equation

Answer 6

i dont understand how you combined that and got a 2x

Answer 7

\[x^2 + (x+1)^2\]

(x+1)^2 = x^2 + 2x +1

therefore \[x^2 + (x+1)^2 = x^2 + x^2 + 2x + 1\]

and then combine the two x^2
\[\implies 2x^2 + 2x +1\]

Answer 8

If I might just point out that these are the sides of the classic 3:4:5 right triangle.

Answer 9

maybe you're not familiar with the shortcut of (x+1)^2 i'll show it slowly

\[(x+1)^2 \implies (x+1)(x+1) \implies x^2 + x + x + 1\]
\[\implies x^2 + 2x +1\]
that's where the 2x was coming from

Answer 10

ohh thank you so so much! you're such a great help :)

Answer 11

you're a great learner too ^_^