root 2xsquare-x+root2
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Hey!!!
=> √2x² + x + √2 = 0
=> √2x² + 2x - x + √2 = 0
=> √2x(x + √2) -(x + √2) = 0
=> (√2x - 1)( x + √2) = 0
Thus x = 1/√2
and x = -√2
=> √2x² + x + √2 = 0
=> √2x² + 2x - x + √2 = 0
=> √2x(x + √2) -(x + √2) = 0
=> (√2x - 1)( x + √2) = 0
Thus x = 1/√2
and x = -√2
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