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

Given:

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

To do:

We have to find the multiples.

Solution:

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

According to the question,

$(7x)^2+(7x+7)^2=637$

$49x^2+49x^2+98x+49=637$

$49(2x^2+2x+1)=49\times13$

$2x^2+2x+1=13$

$2x^2+2x+1-13=0$

$2x^2+2x-12=0$

$2(x^2+x-6)=0$

$x^2+x-6=0$

Solving for $x$ by factorization method, we get,

$x^2+3x-2x-6=0$

$x(x+3)-2(x+3)=0$

$(x+3)(x-2)=0$

$x+3=0$ or $x-2=0$

$x=-3$ or $x=2$

The value of $x$ cannot be negative. Therefore, the value of $x$ is $2$.

$7x=7(2)=14$

$7x+7=14+7=21$

The required multiples are $14$ and $21$.

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