Question

find three consecutive positive integers such that the square of the first plus the third is 8

Answer 1

@MattVictor1990 Hi,
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

Let those 3 consecutive numbers be 'x' , 'x+1' and 'x+2'
square of the first------>x^2
the square of the first plus the third ----->x^2+(x+2)
is 8 -------------> x^2+(x+2) =8
can you solve this quadratic equation ??

Answer 2

Not really need more info because the third is x+4.

Answer 3

the 3rd integer is x+2 actually.
1st integer =x
2nd integer = x+1
3rd integer =x+2

Answer 4

okay thanks

Answer 5

welcome ^_^
ask if any more doubts.