Question

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

Answer 1

Step-by-step explanation:

Let the first number be 'X', so the other number will be'(x+1)'

∵x

2

+(x+1)

2

=421

x

2

+x

2

+1+2x=421

On simplifying further,

2x

2

+2x−420=0

x

2

+x−210=0

On applying Sreedhracharya formula

2a

−b±

b

2

−4ac

2(1)

−(1)±

(1)

2

−4(1)(−210)

2(1)

−1±

1+840

2

−1+

841

2

−1±29

∴X=−15orX=14.

Answer 2

Answer:

Let the two consecutive natural numbers be x and (x + 1) Since, sum of squares of the two consecutive natural numbers = 421 ⇒ x2 + (x + 1)2 = 421 ⇒ x2 + x2 + 2x + 1 – 421 = 0 [ ∵ (a+b)2 = a2 + b2 + 2ab ] ⇒ 2x2 + 2x – 420 = 0 Dividing the above equation by 2 we will get ⇒ x2 + x – 210 = 0 ⇒ x2 + 15x – 14x – 210 = 0 ⇒ x(x + 15) – 14(x+15) = 0 ⇒ (x + 15)(x – 14) = 0 ⇒ x + 15 = 0 or x – 14 = 0 ⇒ x = – 15 or 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.