If the sum of the squares of two consecutive multiples of 7 is 637 then find the multiples
Answers
Hey friend, Here is your answer-
Let the two consecutive multiple of 7 are x and (x+7)
From the given condition,
x^2 + (x+7)^2 = 637
x^2 + x^2 + 14x + 49 = 637
2x^2 +14x +49-637 =0
2x^2 +14x- 588 =0
x^2 + 7x - 294 = 0
x^2 + 21x - 14x - 294 = 0
x ( x + 21) -14 (x+21)=0
(x+21)(x-14)=0
(x+21)=0 or (x-14) =0
x = -21 or x=14
x= -21 is not acceptable.
Therefore x=14
x+7 = 14+7 = 21
The two consecutive multiples of 7 are 14 and 21.
Hope it will help you
Answer:
Let the required consecutive multiples of 7 be 7x and 7 (x+1) .
Let the required consecutive multiples of 7 be 7x and 7 (x+1) .Then,
Let the required consecutive multiples of 7 be 7x and 7 (x+1) .Then, (7x)^ 2 + {7 ( x+1)} ^2 = 637 .
=> 49x^2+ (7x+7)^2 =637
=> 98 x ^2 + 98x - 588= 0
=> x^2 + x - 6 = 0
=> x^2 + 3x - 2x - 6 =0
=> x ( x + 3 ) -2 ( x +3 ) =0
=> ( x + 3 ) ( x - 2 ) = 0
=> x = - 3 or x = 2 .
=> x = 2 ( rejecting x = - 3 )
Hence , the required numbers are ( 7 × 2 ) and (7×3 ) are i.e. , 14 and 21 .
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