Question

The sum of the squares of three consecutive natural numbers is 149. Find the numbers.

Answer 1

Hint: Take the first number to be x, therefore the second and third number becomes (x+1) and (x+2).

Complete step by step answer:
Let, the first number be x,
Therefore, the second number becomes (x+1),
And, the third number becomes (x+2).
Now the condition given is the sum of squares of these three numbers is 149,
So,
\[{x^2} + {\left( {x + 1} \right)^2} + {\left( {x + 2} \right)^2} = 149\]
We know that ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
So, on applying this formula to the above equation, we get
${x^2} + {x^2} + 2x + 1 + {x^2} + 4x + 4 = 149$
$3{x^2} + 6x + 5 = 149$
$3{x^2} + 6x + 5 = 149$
$3{x^2} + 6x - 144 = 0$
${x^2} + 2x - 48 = 0$
Now, let us factorise the equation
$
  {x^2} + 8x - 6x - 48 = 0 \\
  x(x + 8) - 6(x + 8) = 0 \\
  (x + 8)(x - 6) = 0 \\
   \Rightarrow x = 6, - 8 \\
 $
Since, it is given that x is a natural number, so we can write the value of x=6
So,
x=6 is the first number
Therefore,
(x+1) = 6+1= 7 = second number
And (x+2) = 6+2 = 8 = third number

Note: To solve the above equation use the formula to find the roots of a quadratic equation. If the equation cannot be factorised, then apply the formula and find the roots