Question

The sum of the squares of two consecutive multiples of $7$ is $1225$. Find the multiples.

Answer 1

Let the first multiple be $7x$ and next multiple be $7(x+1)$

According to the question

$(7x{)}^2+(7(x+1){)}^2=1225$

$49x^2+49(x^2+2x+1)=1225$

$49(x^2+x^2+2x+1)=1225$

$2x^2+2x+1=\frac{1225}{49}$

$2x^2+2x+1=25$

$2x^2+2x+1-25=0$

$2x^2+2x-24=0$

$x^2+x-12=0$

$x^2+4x-3x-12=0$

$x(x+4)-3(x+4)=0$

$(x+4)(x-3)=0$

$x=3$ or $x=-4$

Since multiples will not be negative so $x=-4$ is rejected

Consecutive multiple of $7$ are $7\times 3=21$ and $7(3+1)=28$

Therefore the two consecutive multiples of $7$ are $21$ and $28$