anonymous
  • anonymous
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 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
random231
  • random231
|dw:1441477010872:dw| now square both sides and solve the eqn
Jhannybean
  • Jhannybean
\[2\sqrt{4}-2x=2-x\] Rearrange.\[2\sqrt{4}-2=2x-x\]\[\boxed{x=2\sqrt{4}-2}\]
IrishBoy123
  • IrishBoy123
\[\sqrt{4}\] is this for real?!

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anonymous
  • anonymous
No, I think it's four real xD im sorry i had to
Jhannybean
  • Jhannybean
You didnt simplify my answer? :(
IrishBoy123
  • IrishBoy123
it's "four real" !!! lol!!
anonymous
  • anonymous
jhannybean, the 2x is also the under the radical. so can we really send it to the other side ?
Jhannybean
  • Jhannybean
so your function is \(2\sqrt{4-2x}= 2-x\)?
anonymous
  • anonymous
yeah
Jhannybean
  • Jhannybean
Okay. Then we would have to begin by squaring both sides. OUr goal is to isolate the x's to one side.
Jhannybean
  • Jhannybean
\[(2\sqrt{4-2x})^2 = (2-x)^2\]\[4(4-2x) = 4-4x+x^2\]\[16-8x=4-4x+x^2\]\[x^2-4x+8x=16-4\]\[x^2+4x-12=0\]Can you factor this now?
anonymous
  • anonymous
um is it (x + 2)(x - 6) = 0 ?
Jhannybean
  • Jhannybean
\[(x+2)(x-6) = x^2 -4x-12 \ne x^2+4x-12\]
anonymous
  • anonymous
oh i think i got it is it (x - 2)( x + 6 ) ?
Jhannybean
  • Jhannybean
Yep
anonymous
  • anonymous
then we simplify like this : x - 2 = 0 x = 2 and x + 6 = 0 x = - 6 so the answer is x = 2, -6
Jhannybean
  • Jhannybean
Mmhmm
anonymous
  • anonymous
omg thank you so much :)
Jhannybean
  • Jhannybean
No problem :)
anonymous
  • anonymous
can i ask u one more question ?
Jhannybean
  • Jhannybean
Sure
anonymous
  • anonymous
it is similar, \[\sqrt{5x - x } = 0\]
anonymous
  • anonymous
how do we solve this ?
anonymous
  • anonymous
Square both sides first, right?
Jhannybean
  • Jhannybean
Again, we want to isolate x so we need to square both sides. Same method.
anonymous
  • anonymous
wouldnt it just be like 4x = 0 at the end? xD or just plain out 0?
Jhannybean
  • Jhannybean
You are correct.
anonymous
  • anonymous
Oooooooo, look at me im doin stuff xD
anonymous
  • anonymous
oh sorry , made a mistake again i meant \[\sqrt{5x} - x = 0 \]
anonymous
  • anonymous
Wouldnt that be the same outcome?
anonymous
  • anonymous
Er... nevermind
Jhannybean
  • Jhannybean
\[\sqrt{5x}-x=0\]\[\sqrt{5x}=x\]\[(\sqrt{5x})^2=x^2\]\[5x=x^2\]\[x^2-5x=0\]\[x(x-5)=0\]\[x=0\]\[x-5=0 \iff x=5\]
Jhannybean
  • Jhannybean
@BweadedChicken no, not the same outccome.
anonymous
  • anonymous
oh that was it ? i thought we need to complete the square and all any way , thank you you're the best :)
Jhannybean
  • Jhannybean
Yeah, completing the square comes along with complex quadratic functions that need to be simplified. You'll know when you see them.
anonymous
  • anonymous
oh ok :)

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