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mdavis
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Squaring Sides 13 Oct 2008, 12:50
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Could someone explain this for me? An answer in the 11th edition of the guide gives this answer for part of the diagnostic test:

square root of 3-2x=square root of 2x +1 (the +1 is not under brackets of the square root)

So it says to square both sides:

(square root of 3-2x)^2=(square root of 2x +1)^2

Then it has this:

3-2x=2x+2(square root of 2x)+1

I am not seeing how they got the terms for the side on the right. It looks like there is an addition +2 and a term that wasn't squared.

Could someone please explain this? If anyone has the 11th edition guide, the answer explanation is on page 53, upper left hand corner of the page.

Thanks in advance!



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scthakur
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Squaring Sides 13 Oct 2008, 13:06
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If the right side is sqrt(2x) + 1 then after squaring it, you will get what is mentioned.
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mdavis
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Squaring Sides 13 Oct 2008, 13:15
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Would you be able to show how that is done? I'm just not seeing it. I understand on the left side of the equation because the root sign is over all the equation.

Would someone be able to break this down for me?
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possible_phd
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Squaring Sides 13 Oct 2008, 21:50
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Use FOIL, just like any other quantities being multiplied together:

[sqrt(2x) + 1] [sqrt(2x) + 1] =

sqrt(2x) * sqrt(2x) + sqrt(2x) + sqrt(2x) + 1 =

2x + 2*sqrt(2x) + 1



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