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

Without doing the algebra, you could consider the smallest Pythagorean triple.

Answer 2

let the smallest number be x
so,other 2 numbers are x+1,x+2
so \(x^2+(x+1)^2=(x+2)^2\)
solve for x.

Answer 3

how?

Answer 4

use \( (a+b)^2=a^2+2ab+b^2 \)

Answer 5

u will get a quadratic in x.

Answer 6

im sorry but can you help me step by step?

Answer 7

\(x^2+(x+1)^2=(x+2)^2-->x^2+x^2+2x+1=x^2+4x+4\)
ok?

Answer 8

which gives
\(x^2-2x-3=0\)
which when u solve gives,x=-1,3
but as x is positive,x=3 only.
so the numbers are 3,4,5.

Answer 9

So next step is \[2x ^{2} +2x+1=x^{2}+4x+2\]

Answer 10

u can subtract x^2+4x+4 from both sides.

Answer 11

but how do i go from
x^2-2x-3=0 to x=3?

Answer 12

thats quadratic equation!
u know any method to solve quadratic equations??

Answer 13

no...

Answer 14

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

Use wolfram or something to solve for it lol.

Answer 15

ok,
x^2-2x-3=0
so, \(x^2-3x+x-3=0\)
\(x(x-3)+1(x-3)=0\)
\((x+1)(x-3)=0\)
ok ?
which step u did not understand?

Answer 16

am9998, if you do not know how to solve quadratic equations, why are you working on this problem? Are you expected to use guess-and-check methods?

Answer 17

No we were expected to know this. but my I didnt learn this.

Answer 18

so far i understand but whats the next step?

Answer 19

\((x+1)(x-3)=0\)
so, x+1=0 OR x-3=0
so, x=-1 OR x=3
but x is positive
so x=3 only.

Answer 20

got it?

Answer 21

yes thanks

Answer 22

welcome :)