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