Question

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

Answer 1

Let the required consecutive multiplies of 7 be 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=0  or  x-3=0` 

⇒`x=-4  or  x=3` 

∴x=3            (Neglecting the negative value) 

When x=3,

`7x=7xx3=21` 

`7(x+1)=7(3+1)=7xx4=28` 

Hence, the required multiples are 21 and 28.