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Solve. Please show your work. √(2x+1)=x-1
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\[\sqrt{2x+1}=x-1\]
\[ \sqrt{2x+1}=x-1\] Take a square on both sides,. \[ (\sqrt{2x+1})^2=(x-1)^2\] this will help in getting rid of the square root, \[2x+1 = (x-1)^2\] using FOIL to solve the right side. \[2x+1=x^2-2x+1\]\[x^2+4x=0\] after solving u will get, x=4 and x=0
now use them to check in the original equation, insert 0 first √(2x+1)=x-1 sqrt(2(0)+1) = 0-1 sqrt 1 = -1 so zero is not a solution, next check with 4. insert 4 √(2x+1)=x-1 sqrt(2(4)+1) = 4-1 sqrt9 = 3 3 = 3 So this proves that ONLY FOUR is your solution/answer.
i hope that will help u. =)
lil, do u get it as well?
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