Mark M. answered • 03/12/16
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√2 x2 - 3x + √2 = 0
√2 x2 + √2 = 3x
√2 (x2 + 1) = 3x
2(x2 + 1)2 = 9x2
2(x4 + 2x2 + 1) = 9x2
2x4 + 4x2 + 2 = 9x2
2x4 - 5x2 + 2 = 0
(2x2 - 1)(x2 - 2) = 0
2x2 = 1
x2 = 1/2
x = ±√2 / 2
x2 = 2
x = ±√2
Mark M.
So to reposition Gauss, present your solution!
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03/12/16
Kenneth S.
I sent the comment (mentioning Gauss) immediately after I sent in my solution.
My solution was verified decimally (albeit not included in my posting); I have also solved the given quadratic using the Quadratic formula (albeit not in my posting). I
f you click on this problem, you ought to have seen my solution immediately following yours. Two of your solutions are extraneous, because a quadratic has 2 (complex) zeros.
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03/13/16
Kenneth S.
03/12/16