Grade 8Algebra
find two natural numbers whose difference is 4 and sum of their squares is 8
1 Answer
Santosh Kumar Thakur
It is not possible. Because
Let the two numbers be A and B where A>B.
Now, A/Q
A^2 + B^2 =8 ….....(i)
A – B = 4 …......(ii)
From eqn. (ii) we get
A = 4 +B
Now putting the value of A in eqn (i) we get
(4+B)^2 + B^2 = 8
=> 16 + B^2 + 8B + B^2 = 8
=> 2B^2 + 8B + 8 = 0
So, B = -2 (By Quadratic Equation)
But -2 is not a natural number.
Hence It isn.t possible.
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