anonymous
  • anonymous
(squaredrootof)x-5=(squaredrootof)x-1 [the -1 is not inside the squaredroot) :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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dumbcow
  • dumbcow
square both sides you have to FOIL the right side
anonymous
  • anonymous
uhhh.... i need help with that :///
anonymous
  • anonymous
\[\sqrt{x-5}=\sqrt{x}-1\]\[(\sqrt{x-5})^2=(\sqrt{x}-1)^2\]v-------------^\[(\sqrt{x}-1)(\sqrt{x}-1)\]FOIL \[(x-\sqrt{x}-\sqrt{x}+1)\]\[(x-2\sqrt{x}-1)\]\[x-5=x-2\sqrt{x}+1\]-x -1 -x -1 \[-6=-2\sqrt{x}\]/-2 /-2\[3=\sqrt{x}\]\[(3)^2=(\sqrt{x})^2\]\[9=x\]Finished!!!

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anonymous
  • anonymous
woah.. id nt get the foil.. why woudl (sqrdrt) of x multiplied by the same thing give x?
anonymous
  • anonymous
I forget the name of the actual rule. But simply it's\[(x^2)(x^3)=x^5\]What you do is take the exponents and add them together.\[\sqrt{x}=x ^{1/2}\]Therefore\[(\sqrt{x})(\sqrt{x})=(x ^{1/2})(x ^{1/2})\]Which then makes\[(x ^{1/2})(x ^{1/2})=x^1\]or x.
anonymous
  • anonymous
oh. haha. sorry :) and i don't get the x-1-x-1 part.. :(
anonymous
  • anonymous
\[(x ^{a})(x ^{b})=x ^{a+b}\] That's the rule.
anonymous
  • anonymous
I'm subtracting x and 1 from each side.
anonymous
  • anonymous
ooohh... i see it now:) can you repeart that last part? after the little isue i just did? my computer won't show anything after that. :(
anonymous
  • anonymous
after the -x-1-x-1?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
\[x-5=x-2\sqrt{x}+1\]-x -1 -x -1\[-6=-2\sqrt{x}\]/-2 /-2\[3=\sqrt{x}\]\[(3)^2=(\sqrt{x})^2\]\[9=x\]
anonymous
  • anonymous
what do the /-2/-2 stand for?
anonymous
  • anonymous
Divide by -2 on both sides.
anonymous
  • anonymous
okay. :)
anonymous
  • anonymous
I was taught "The Granny Rule: Granny's are fair, what they do to one chile they do to the other." Same with algebra, what you do to one side of the equation you have to do to the other.
anonymous
  • anonymous
*child*
anonymous
  • anonymous
mahaha:) that's a good way to remember it:D

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