How many pairs of natural bombers are there such that the difference of whose squares is 63?
Answers
Answer:
3 pairs . they are (32,31) , (12,9) & (8,1)
Step-by-step explanation:
(x² - y²) = 63
Or, (x - y)(x + y) = 63
1 x 63 = 63
3 x 21 = 63
7 x 9 = 63
Thus, the factors of 63 possibles are 1, 3, 7, 9, 21 and 63
x+y = 63 x+y =21 x+y = 9
x-y = 1 x-y = 3 x-y = 7
______ ______ ______
2x = 64 2x = 24 2x = 16
x = 32 x = 12 x = 8
=>
y = 31 y = 9 y = 1 (on subst. x value in any of the eqn.s)
so 3 pairs exists, (32,31) , (12,9) & (8,1)
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verify:
(32+31)(32-31) = 63
=> 63*1 = 63
=> 63=63
& (12+9)(12-9) = 63
=> 21*3 =63
=> 63=63
then (8+1)(8-1) = 63
9*7 = 63
63 = 63
Hence verified
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Answer:
No. of pairs of natural numbers such that the difference of their squares is 63 = 3
Step-by-step explanation:
Recall the formula,
x² - y² = (x+y)(x-y)
Let 'x' and 'y' be the two numbers
Given that the difference of their squares = 63
That is, x² - y² = 63
(x+y)(x-y) = 63 ------------(A)
Now, the factors of 63 are 1,3,7,9,21,63
The number of ways equation (A) can be written as
- (x+y)(x-y) = 1 ×63
- (x+y)(x-y) = 3 × 21
- (x+y)(x-y) = 7×9
Then, we get six pairs of equations as given below
- x+y = 1 and x- y = 63
- x+ y = 63 and x -y = 1
- x+y = 3 and x- y = 21
- x+y = 21 and x -y = 3
- x+y = 7 and x -y = 9
- x+y = 9 and x -y = 7
Solving these six pairs of equations we get,
- There is no two natural numbers such that x+y =1
- Solving x+ y = 63 and x -y = 1 we get, x = 32 and y = 31
- Solving x+y = 3 and x- y = 21 we get, x = 12 and y = -9
- Solving x+y = 21 and x -y = 3 we get, x = 12 and y = 9
- Solving x+y = 7 and x -y = 9 we get, x = 8 and y = -1
- Solving x+y = 9 and x -y = 7 we get, x = 8 and y = 1
Hence, Out of these 6 pairs of solutions, three pairs of solutions are natural numbers.
∴No. of pairs of natural numbers such that the difference of their squares is 63 = 3
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