#### Question

The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.

#### Solution

Let one of the number be 7x then the other number be 7(x + 1).

Then according to question,

\[\left( 7x \right)^2 + \left[ 7\left( x + 1 \right) \right]^2 = 637\]

\[ \Rightarrow 49 x^2 + 49( x^2 + 2x + 1) = 637\]

\[ \Rightarrow 49 x^2 + 49 x^2 + 98x + 49 - 637 = 0\]

\[ \Rightarrow 98 x^2 + 98x - 588 = 0\]

\[ \Rightarrow x^2 + x - 6 = 0\]

\[ \Rightarrow x^2 + 3x - 2x - 6 = 0\]

\[ \Rightarrow x(x + 3) - 2(x + 3) = 0\]

\[ \Rightarrow (x - 2)(x + 3) = 0\]

\[ \Rightarrow x - 2 = 0 \text { or } x + 3 = 0\]

\[ \Rightarrow x = 2 \text { or } x = - 3\]

Since, the numbers are multiples of 7,

Therefore, one number = 7 × 2 =14.

Then another number will be \[7(x + 1) = 7 \times 3 = 21\]

Thus, the two consecutive multiples of 7 are 14 and 21.