Prove that ( 3-root7 ) holds square is an irrational number
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LET 3-ROOT 7 BE EQUAL TO X THAT IS,
3-√7 = X
Squaring on both side...we get,
(3-√7)*2 = X*2
9 +7 + 6√7 = X*2
16 + 6√7 = X*2
THEN,
(X*2 - 16)/6 = √7
NOW AS √7 IS AN IRRATIONAL NUMBER , IT IS PROVED THAT (3-√7)SQUARE
HOLDS AN IRRATIONAL NUMBER.
3-√7 = X
Squaring on both side...we get,
(3-√7)*2 = X*2
9 +7 + 6√7 = X*2
16 + 6√7 = X*2
THEN,
(X*2 - 16)/6 = √7
NOW AS √7 IS AN IRRATIONAL NUMBER , IT IS PROVED THAT (3-√7)SQUARE
HOLDS AN IRRATIONAL NUMBER.
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