find the zeroes of 2s2 -(1+2root2)s+root2
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2s^2-s-2√2s+√2 s(2s-1) -√2(2s-1) (s-√2) (2s-1) S=√2 ,s=1/2
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Hey,!!!
We have
2s² - (1 + 2√2)s + √2 = 0
=> 2s² - s - 2√2s + √2 = 0
=> s(2s - 1) - √2(2s - 1) = 0
=> (s - √2)(2s - 1) =0
Thus => s - √2 = 0
=> s = √2
and 2s - 1 = 0
=> s = 1/2
Hope this helps
We have
2s² - (1 + 2√2)s + √2 = 0
=> 2s² - s - 2√2s + √2 = 0
=> s(2s - 1) - √2(2s - 1) = 0
=> (s - √2)(2s - 1) =0
Thus => s - √2 = 0
=> s = √2
and 2s - 1 = 0
=> s = 1/2
Hope this helps
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