please help !!!!! Choose any four (4) of the five radical equations below. Solve each equation, showing all steps leading to your answer. Then check your answer to determine if any of the solutions are extraneous i just need you guys to help me in one question 4. 2√4 - 2x = 2 - x please help !!! how do we solve it ?

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please help !!!!! Choose any four (4) of the five radical equations below. Solve each equation, showing all steps leading to your answer. Then check your answer to determine if any of the solutions are extraneous i just need you guys to help me in one question 4. 2√4 - 2x = 2 - x please help !!! how do we solve it ?

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|dw:1441477010872:dw| now square both sides and solve the eqn
\[2\sqrt{4}-2x=2-x\] Rearrange.\[2\sqrt{4}-2=2x-x\]\[\boxed{x=2\sqrt{4}-2}\]
\[\sqrt{4}\] is this for real?!

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No, I think it's four real xD im sorry i had to
You didnt simplify my answer? :(
it's "four real" !!! lol!!
jhannybean, the 2x is also the under the radical. so can we really send it to the other side ?
so your function is \(2\sqrt{4-2x}= 2-x\)?
yeah
Okay. Then we would have to begin by squaring both sides. OUr goal is to isolate the x's to one side.
\[(2\sqrt{4-2x})^2 = (2-x)^2\]\[4(4-2x) = 4-4x+x^2\]\[16-8x=4-4x+x^2\]\[x^2-4x+8x=16-4\]\[x^2+4x-12=0\]Can you factor this now?
um is it (x + 2)(x - 6) = 0 ?
\[(x+2)(x-6) = x^2 -4x-12 \ne x^2+4x-12\]
oh i think i got it is it (x - 2)( x + 6 ) ?
Yep
then we simplify like this : x - 2 = 0 x = 2 and x + 6 = 0 x = - 6 so the answer is x = 2, -6
Mmhmm
omg thank you so much :)
No problem :)
can i ask u one more question ?
Sure
it is similar, \[\sqrt{5x - x } = 0\]
how do we solve this ?
Square both sides first, right?
Again, we want to isolate x so we need to square both sides. Same method.
wouldnt it just be like 4x = 0 at the end? xD or just plain out 0?
You are correct.
Oooooooo, look at me im doin stuff xD
oh sorry , made a mistake again i meant \[\sqrt{5x} - x = 0 \]
Wouldnt that be the same outcome?
Er... nevermind
\[\sqrt{5x}-x=0\]\[\sqrt{5x}=x\]\[(\sqrt{5x})^2=x^2\]\[5x=x^2\]\[x^2-5x=0\]\[x(x-5)=0\]\[x=0\]\[x-5=0 \iff x=5\]
@BweadedChicken no, not the same outccome.
oh that was it ? i thought we need to complete the square and all any way , thank you you're the best :)
Yeah, completing the square comes along with complex quadratic functions that need to be simplified. You'll know when you see them.
oh ok :)

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