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OG 12 Diagnostic Square Root Question D17

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OG 12 Diagnostic Square Root Question D17

by EMAN » Thu Sep 10, 2009 6:23 pm
If (Square Root of 3 - 2x) = (Square Root of 2x) + 1, then 4x^2 =

A - 1
B - 4
C - 2 - 2x
D - 4x - 2
E - 6x - 1

The explanation in the OG has about eight steps of simplifying and is thoroughly confusing for someone who does not have a strong math background. Can someone please let me know if there is a relatively simple way to solve this equation?? Any help is greatly appreciated.
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by Nermal » Fri Sep 11, 2009 7:55 am
IMO E.

First and most importantly we must get rid of the squareroots.
There is only one way: raise both sides to the power of 2.
sqroot(3-2x)=(sqroot2x)+1
on the right side it is the first binomial theorem.
[It is important that you know those three by heart!]

3-2x=2x+2(sqroot(2x))+1

Now you must bring everything except of the squareroot to one side. This way the next time we square we will get rid of this squareroot.

2-4x=2(sqroot(2x)). Now we square both sides.
(Now it is the second binomial theorem on the left side)
4-2*2*4x+16x^2=4*2x
4-16x+16x^2=8x
16x^2-24x+4=0
Divide everything by 4:
4x^2-6x+1=0
4x^2=6x-1

Since we look for 4x^2 we are done now.
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Follow Up

by EMAN » Fri Sep 11, 2009 8:58 am
Great job Nermal. E is correct. Can you tell how you derived the equation below once you squared both sides?

3-2x=2x+2(sqroot(2x))+1

Are you just taking the square root of 4x^2 and then getting 2x? If that's the case, this makes sense but I guess I was initially confused why one has to put the 4x^2 on the right side of the equation instead of the left when framing this problem. Once I'm clear on that I think I'm golden.
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Re: Follow Up

by Nermal » Fri Sep 11, 2009 10:35 am
Great job Nermal. E is correct. Can you tell how you derived the equation below once you squared both sides?

3-2x=2x+2(sqroot(2x))+1
Left side:
I guess you know that by squaring a squareroot you just get what before was under the squareroot sign:
sqroot(3-2x)*sqroot(3-2x)=3+2x

Right side of the equation (a little more complicated):
Use the binomial theorem no 1 which is in its generell form as follows:
(x+y)^2=(x+y)*(x+y)=x^2+2xy+y^2

here: x=(sqroot2x), y=1
plug it in and you get:
(sqroot(2x)+1)^2=(sqroot(2x)+1)*(sqroot(2x)+1)=
2x+2*(sqroot(2x))*1+1*1=2x+2(sqroot(2x))+1

Now get the term with the squareroot on one side.
It's the only way of getting rid of the squareroot.
3-2x=2x+2(sqroot(2x))+1
3-1-2x-2x=2(sqroot(2x))
2-4x=2(sqroot(2x))

Now square both sides of the equation again.

Left side (use second binomial theorem; in generell form: (x-y)^2=(x-y)*(x-y)=x^2-2xy+y^2

here: x=2; y=4x
2^2-2*2*4x+(4x)^2=4-16x+16x^2

right side: (2(sqroot(2x)))^2=4*2x=8x

Take both sides of the equation together:
4-16x+16x^2=8x

substract 4 from and add 16x to both sides and divide everything by 4:
4x^2=6x-1

HTH
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Got it

by EMAN » Fri Sep 11, 2009 4:46 pm
Okay I got it now. I had an issue trying to do the binomial theorem and became confused with the radical signs. I got confused thinking that the SQRT 2x * SQRT 2x equaled the SQRT of 4x instead of just 2x. Additionally, adding SQRT of 2x to the SQRT of 2x threw me off at first too. I would have wrote it as the SQRT of 4x instead 2 * SQRT of 2x (which appears lessing confusing to me at least). Thanks!
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by Nermal » Sat Sep 12, 2009 1:13 am
I think it becomes clear when thinking of sqroot 2
as 2^(1/2)

then you see that sqroot2*sqroot2=2^(1/2)*2^(1/2)
Now you have exponents with the same base so in order to multiply them you have to keep the base and add the exponents:
2^((1/2)+(1/2))=2^1=2
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