Question

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

Answer 1

Let the two consecutive natural numbers be x and (x + 1)

Given x2 + (x + 1)2
= 421 x2 + x2 + 2x + 1 – 421
= 0 2x2 + 2x – 420 = 0 x2 + x – 210
= 0 x2 + 15x – 14x – 210
= 0 (x + 15)(x – 14) = 0
Hence x = -15 or 14
Since x is a natural number
Thus x = 14 x + 1 = 15
Therefore the two consecutive numbers are 14 and 15.

Answer 2

Answer:14 and 15

Explaination:

Let frist consecutive number be x

and second consecutive number be(x+1)

According to the question sum of the square of two consecutive natural number is 421

so,. (x)²+(x+1)²=421

=x²+x²+2x1+1²=421

=x²+x²+2x+1=421

=2x²+2x=421-1

=2x²+2x=420

=2x²+2x-420=0

divide both side by 2

=2x²/2+2x/2-420/2

=x²+x-210

=x²+ 15x – 14x – 210 = 0

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

(x+15) (x-14)

x=-15 and x=14

Since x is a natural number x can not be negative So x = 14 And (x + 1) = 14+1 = 15 Therefore, the two consecutive numbers are 14 and 15.