Question

3.The sum of the squares of the two consecutive
multiples of 7 is 637, find the multiples?​

Answer 1

Answer:

Let the multiples of 7x&7(x+1)

(7x)2+(7(x+1))2=637

⇒49x2+49(x+1)2=637

⇒x2+(x+1)2=13

⇒x2+x2+1+2x−13=0

⇒2x2+2x−12=0

⇒x2+x−6=0

⇒(x+3)(x−2)=0

so x=-3 or x=2

no s= -14,-21 no s=14,21

Answer 2

Let one of the multiple be x
Therefore the other can be x+7

we take x to be 14 and reject the -21
therefore x+7 = 21

ANSWER WITH STEPS ⬇️