Question

find two consecutive natural numbers, the sum of whose square is 145

Answer 1

Its ur answer Bro.


Let one number be x
Then other number = x+1



It i s given that the sum of square of those numbers is 145.

Therefore, 
x² + (x+1)² = 145.
x² + x² + 2x + 1 = 145              {(a+b)²= a² + 2ab + b²}
2x² + 2x - 144 = 0 

Dividing both sides by 2, we obtain
x² + x - 72 = 0
x² - 8x + 9x - 72 = 0
x(x-8) + 9 (x-8) = 0
(x+9) (x-8) = 0
x = -9  or x=8 


As x = -9 can't be answer because it is not a natural number.

Hence, x = 8.
And, x+1 = 9.



Justification:-
                         8² + 9²
                       = 64 + 81 = 145

Hence, it is proved.




#RehanAhmadXLX
#BrainlyStar

Answer 2

Step-by-step explanation:

let,

first number = x

second number= x+1

given that,

                 sum of whose squares=145

                 $x^2$+$(x+1)^2$=145

                  $x^2$+$x^2$+$1^2$+2(x)(1)=145

                 2$x^2$+1+2x=145

                 2$x^2$+2x=145-1

                 2$x^2$+2x=144

                 2$x^2$+2x-144=0

 here, 2 is common then,

                 $x^2$+x-72=0

splitting the middle term,

              $x^2$-8x+9x-72=0

            x(x-8)+9(x-8)=0

              (x-8)(x+9)=0

              (x-8)=0 or (x+9)=0

              x=8 or x= -9

verification:

$(8)^2$+$(9)^2$=145

64+81=145

145=145

hence proved.