Question

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

Answer 1

Let the required consecutive multiples of 7 are 7x and 7(x + 1).

According to the given condition,

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

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

$49x^2+49x^2+98x+49=1225$

$98x^2+98x-1176=0$

$x^2+x-12=0$

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

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

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

$x+4=0orx-3=0$

$x=-4orx=3$

Therefore, $x=3$(Neglecting the negative value)

When $x=3,$

$7x=7\times 3=21$

$7(x+1)=7(3+1)=7\times 4=28$

Hence, the required multiples are 21 and 28.