The sum of square of two consecutive multiple of 7 is 637 find the multiples.
Answers
Here is your answer:-
Let one number be x.
Second number=x+7. (both are the multiples of 7, so they must have a difference of 7).
x²+(x+7)²=637.
x²+x²+2×x×7+49=637.
2x²+14x-588=0.
2(x²+7x-294)=0.
x²+7x-294=0/2.
x²+7x-294=0.
Now, we will solve it by Middle Term Splitting Method;
x²+21x-14x-294=0.
x(x+21)-14(x+21)=0.
(x-14)(x+21)=0.
x=+14,-21.
Hence the first number is+14 or -21.
So, the second number = 14+7 or -21+7.
= 21 or -14.
Hence the two numbers will be 14 & 21 or -14 & -21.
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Answer:
Step-by-step explanation:
Solution :-
Let the 1st required consecutive multiple of 7 be 7x
And the 2nd consecutive multiple of 7 be 7(x + 1).
Then,
According to the Question,
⇒ (7x)² + [7(x + 1)²] = 637
⇒ 49x² + (7x + 7)² = 637
⇒ 98x² + 98x - 588 = 0
⇒ x² + x - 6 = 0
⇒ x² + 3x - 2x - 6 = 0
⇒ x(x + 3) - 2(x + 3) = 0
⇒ (x + 3) (x - 2) = 0
⇒ x + 3 = 0 or x - 2 = 0
⇒ x = - 3, 2 (As x can't be negative)
⇒ x = 2
1st Number = 7 × 2 = 14
2nd Number = 7 × 3 = 21
Hence, the required numbers are 14 and 21.