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Mathematics 28 Online
OpenStudy (anonymous):

Solve the equation. Check both solutions and only write the real solution(s). v2x + 3 = x *v stands for sqrt A. {-1} B. {3} C. {-1, 3}

OpenStudy (anonymous):

I think B.

OpenStudy (eyust707):

i think so too! negitive square roots are not real.

OpenStudy (anonymous):

I can't really tell what's going on the way you have the math written. Is it: \[\sqrt{2}x+3=x\] \[\sqrt{2x}+3=x\] or \[\sqrt{2x+3}=x\]

OpenStudy (eyust707):

thats a good point because 3 isnt actually a solution the way u wrote it

OpenStudy (anonymous):

@TheFigure the last one you wrote is how it looks

OpenStudy (anonymous):

Alright - this one. \[\sqrt{2x+3}=x\]

OpenStudy (anonymous):

First, I'd square both sides to get the x out from under that square root:\[\sqrt{2x+3}^2=x^2\]Then, I'd recognize that it is quadratic and move everything to one side. \[0=x^2-2x+3\]From here, you can either use the quadratic formula or factor (depending on where you are in your math class).

OpenStudy (anonymous):

Yikes - I blew it! Let's try that again... It should be:\[0=x^2-2x-3\]

OpenStudy (anonymous):

Now it factors: \[0=x^2-2x-3\] \[0=(x-3)(x+1)\] x=-1, 3

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