Solve / ? 4y + 1 - y = 1
- Thread starter Panama
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Re: Solve
Is the problem
\(\displaystyle \sqrt{4y+1} \, - \, y \, = \, 1\).............should be posted as ?(4y+1) - y =1
or
\(\displaystyle \sqrt{4y \, + \, 1 \, - \, y} \, = \, 1\)..........should be posted as ?(4y+1-y) = 1
or
\(\displaystyle \sqrt{4y} \, + \, 1 \, - \, y \, = \, 1\)..........should be posted as ?(4y)+1-y = 1
Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
Panama said:
Is the problem
\(\displaystyle \sqrt{4y+1} \, - \, y \, = \, 1\).............should be posted as ?(4y+1) - y =1
or
\(\displaystyle \sqrt{4y \, + \, 1 \, - \, y} \, = \, 1\)..........should be posted as ?(4y+1-y) = 1
or
\(\displaystyle \sqrt{4y} \, + \, 1 \, - \, y \, = \, 1\)..........should be posted as ?(4y)+1-y = 1
Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
Re: Solve
Hello, Panama!
The only way it makes sense is: .\(\displaystyle \sqrt{4y-1} - y \:=\:1\)
Isolate the radical: .\(\displaystyle \sqrt{4y-1} \:=\:y + 1\)
Square both sides: .\(\displaystyle \left(\sqrt{4y-1}\right)^2 \:=\
y + 1)^2\)
And we have: .\(\displaystyle 4y - 1 \:=\:y^2 + 2y + 1 \quad\Rightarrow\quad y^2-2y \:=\:0\)
Now solve the equation . . . and watch for extraneous roots.
Hello, Panama!
The only way it makes sense is: .\(\displaystyle \sqrt{4y-1} - y \:=\:1\)
Isolate the radical: .\(\displaystyle \sqrt{4y-1} \:=\:y + 1\)
Square both sides: .\(\displaystyle \left(\sqrt{4y-1}\right)^2 \:=\
y + 1)^2\)And we have: .\(\displaystyle 4y - 1 \:=\:y^2 + 2y + 1 \quad\Rightarrow\quad y^2-2y \:=\:0\)
Now solve the equation . . . and watch for extraneous roots.