Squaring both sides we get y-3 = 1 - 2√(2y-4) + (2y-4)
Isolate the remaining radical: 2√(2y-4) = y
Square both sides: 4(2y-4) = y2
y2 - 8y + 16 = 0
(y-4)2 = 0
y = 4
Check in the original equation
Bianca J.
asked • 10/28/15Solve equation using algebra
√(y-3) = 1 - √(2y-4)
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