Question

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

Answer 1

SOLUTION :  

Let the two consecutive multiples of 7 be x and x +7.

A.T.Q

x² + (x + 7)² = 637

x² + x² + 7² + 2×7x = 637

x² + x² + 49 + 14x = 637

2x² + 14x + 49 - 637 = 0

2x² + 14x - 558 = 0

2(x² + 7x - 294 ) = 0

x² + 7x - 294  = 0

x² + 21x - 14x  - 294  = 0

[By middle term splitting]

x(x + 21) - 14(x + 21) = 0

(x - 14) (x + 21) = 0

(x - 14) or  (x + 21) = 0

x = 14 or x = -21

Since , x can't be negative , so x ≠ - 21

Therefore , x = 14

First Multiple = x = 14  

Second multiple of 7 = x + 7 = 14 + 7 = 21  

Hence, the two consecutive multiples of 7 are 14 & 21 .

HOPE THIS  ANSWER WILL HELP YOU..

Answer 2

Heyaa....☺☺☺

Let the consecutive multiples be x and x+7

$= > {x}^{2} + ({x + 7})^{2} = 637 \\ \\ = > {x}^{2} + {x}^{2} + 49 + 14x = 637 \\ \\ = > 2 {x}^{2} + 14x = 637 - 49 \\ \\ = > 2 {x}^{2} + 14x - 588 = 0 \\ \\ = > 2( {x}^{2} + 7x - 294) = 0 \\ \\ = > now \: by \: middle \: term \: spliting \: method \\ \\ = > {x }^{2} + 21x - 14x - 294 \\ \\ = > x(21 + x) - 14( x + 21) = 0 \\ \\ = > ( x - 14)(x + 21) = 0 \\ \\ = > x = 14 \: or \: x = - 21$

As x cant be negative so the value of x is 14

So x = 14
Then x + 7 = 14 + 7 = 21

✔So the two consecutive multiples of 7 = 14 and 21

Thanks....☺☺☺