Question

The sum of square of three consecutive integers is 110. Find the numbers

Answer 1

Answer:

Step-by-step explanation:

Let three consecutive natural number be X, X+1, X+2 Then according to problem (x)² + (x+1)²+ (x+2)²=110 x²+x²+1+2x+x²+4+4x-110=0 3x²+6x-105=0 x²+2x-35=0 x²+7x-5x-35=0 x(x+7)-5(x+7)=0 (x+7)(x+5)=0 x=-7 or x=5 But x=-7 is reject as it is not a number Then x=5 Hence, required number are 5,(5+1),(5+2)

Answer 2

Answer:

The required numbers are -7,-6,-5 or 5,6,7

Step-by-step explanation:

Let the numbers be x-1,x,x+1.

Now from the question we obtain $(x-1)^{2} +x^{2} +(x+1)^{2} = 110$

which will give us $x^{2} +x^{2} +x^{2} +2 = 110$

on simplifying we obtain $x^{2} =36$ implies x = ±6.

So the numbers are -7,-6,-5 or 5,6,7