Question

the sum of the square of two consecutive natural number is 421 find the number​

Answer 1

Let the 2 concecutive no.s be x and x+1

Sum of their squares is 421 (given)

i.e., x² + (x+1)² = 421

x²+x²+2x+1=421

2x²+2x=421-1

2(x²+x)=420

x²+x=210

x²+x-210=0

x²+15x-14x+210=0

x(x+15) -14(x+15)

(x+15)(x-14)=0

x+15=0    or     x-14=9

x= -15  or  x=14

-15 is not a natural number

So, First number  x=14

 and second number = x+1 = 14+1 =15

Hence the 2 natural numbers are 14,15

Answer 2

Solution :

Let's 1st number = x

2nd number = x + 1

According to question :

x² + (x+1)² = 421

x² + x² + 2x + 1 = 421

2x² + 2x + 1 - 421 = 0

2x² + 2x - 420 = 0

x² + x - 210 = 0

x² + 15x - 14x - 210 = 0

x ( x + 15 ) -14 ( x + 15 ) = 0

( x - 14 ) ( x + 15 )

x - 14 = 0

x = 14

x + 15 = 0

x = - 15

-15 is not a natural number so value of x is 14

1st number = 14

2nd number = 14 + 1 = 15