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# If root{3-2x} = root(2x) +1, then 4x^2 =

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If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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09 Jul 2012, 03:23
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If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Diagnostic Test
Question: 16
Page: 22
Difficulty: 600
[Reveal] Spoiler: OA

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Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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09 Jul 2012, 03:23
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SOLUTION

If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

$$\sqrt{3-2x} = \sqrt{2x} +1$$ --> square both sides: $$(\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2$$ --> $$3-2x=2x+2*\sqrt{2x}+1$$ --> rearrange so that to have root at one side: $$2-4x=2*\sqrt{2x}$$ --> reduce by 2: $$1-2x=\sqrt{2x}$$ --> square again: $$(1-2x)^2=(\sqrt{2x})^2$$ --> $$1-4x+4x^2=2x$$ --> rearrange again: $$4x^2=6x-1$$.

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Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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09 Jul 2012, 06:16
1
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Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Hi,

Put x=1,
LHS= 1,
RHS=2.414
or LHS < RHS

Put x=0,
LHS= 1.732
RHS=1
LHS>RHS,

Thus, 0< x < 1, for equality to hold true.
or 0 < 2x < 2, squaring,
$$0 < 4x^2 < 4$$

Lets check the options now,
(A) 1 ($$0 < 4x^2 < 4$$)
(B) 4 ($$0 < 4x^2 < 4$$)
(C)2 − 2x (0< x < 1, thus, 0< 2 − 2x < 2, but $$0 < 4x^2 < 4$$)
(D) 4x − 2 (0< x < 1, thus, -2< 4x-2 < 2, but $$0 < 4x^2 < 4$$)
(E) 6x − 1 (0< x < 1, thus, -1< 6x-1 < 5, and $$0 < 4x^2 < 4$$)

Thus, only possible option which lies in the range is E.

Regards,
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Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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09 Jul 2012, 07:26
Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Hi,

Put x=1,
LHS= 1,
RHS=2.414
or LHS < RHS

Put x=0,
LHS= 1.732
RHS=1
LHS>RHS,

Thus, 0< x < 1, for equality to hold true.
or 0 < 2x < 2, squaring,
$$0 < 4x^2 < 4$$

Lets check the options now,
(A) 1 ($$0 < 4x^2 < 4$$)
(B) 4 ($$0 < 4x^2 < 4$$)
(C)2 − 2x (0< x < 1, thus, 0< 2 − 2x < 2, but $$0 < 4x^2 < 4$$)
(D) 4x − 2 (0< x < 1, thus, -2< 4x-2 < 2, but $$0 < 4x^2 < 4$$)
(E) 6x − 1 (0< x < 1, thus, -1< 6x-1 < 5, and $$0 < 4x^2 < 4$$)

Thus, only possible option which lies in the range is E.

Regards,

Hi

Can we solve equations and get the answer?
Like taking $$\sqrt{2x}$$ to right and squaring the both sides
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Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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09 Jul 2012, 21:48
Hi narangvaibhav,

Yes definitely, we can use the conventional method too,
$$\sqrt{3-2x} = \sqrt{2x} +1$$
Squaring both the sides,
$$3-2x=2x+1+2 \sqrt{2x}$$
or $$1-2x=\sqrt{2x}$$
Again, squaring both the sides,
$$1+4x^2-4x=2x$$
or $$4x^2=6x-1$$

Regards,

PS: But that isn't much fun.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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13 Jul 2012, 02:42
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SOLUTION

If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

$$\sqrt{3-2x} = \sqrt{2x} +1$$ --> square both sides: $$(\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2$$ --> $$3-2x=2x+2*\sqrt{2x}+1$$ --> rearrange so that to have root at one side: $$2-4x=2*\sqrt{2x}$$ --> reduce by 2: $$1-2x=\sqrt{2x}$$ --> square again: $$(1-2x)^2=(\sqrt{2x})^2$$ --> $$1-4x+4x^2=2x$$ --> rearrange again: $$4x^2=6x-1$$.

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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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29 Apr 2013, 06:40
Bunuel wrote:
If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Can someone help me solve this?

$$\sqrt{(3-2x)}$$ = $$\sqrt{2x}$$ + 1
$$\sqrt{(3-2x)}$$ + 1 = $$\sqrt{2x}$$
If I square both sides,
2 - $$\sqrt{(3-2x)}$$ = 2x
So, $$4x^2$$ = 4 - (3-2x) - 2$$\sqrt{(3-2x)}$$

I am unable to reach to E

However, if I squared the main problem without shifting 1, then I am reaching to E

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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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29 Apr 2013, 06:56
1
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Expert's post
sdas wrote:
Bunuel wrote:
If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Can someone help me solve this?

$$\sqrt{(3-2x)}$$ = $$\sqrt{2x}$$ + 1
$$\sqrt{(3-2x)}$$ + 1 = $$\sqrt{2x}$$
If I square both sides,
2 - $$\sqrt{(3-2x)}$$ = 2x
So, $$4x^2$$ = 4 - (3-2x) - 2$$\sqrt{(3-2x)}$$

I am unable to reach to E

However, if I squared the main problem without shifting 1, then I am reaching to E

You could solve this way too:

$$\sqrt{3-2x} = \sqrt{2x} +1$$

Re-arrange: $$\sqrt{3-2x}-1 = \sqrt{2x}$$;

Square: $$(3-2x)-2\sqrt{3-2x}+1=2x$$;

Re-arrange and reduce by 2: $$\sqrt{3-2x}=2-2x$$;

Square: $$3-2x=4-8x+4x^2$$

$$4x^2=6x-1$$.

Hope it's clear.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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29 Apr 2013, 08:15
Thanks Bunuel. I missed the last 2 steps....
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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15 Dec 2013, 11:46
Hi guys,

can someone please explain why (root(2x) +1)^2 is not 2x +1 but 2x+1+2 root(2x). I thought I would just take every term to the secound power, such as root(2x)^2 + 1^2...

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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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16 Dec 2013, 01:18
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Expert's post
steilbergauf wrote:
Hi guys,

can someone please explain why (root(2x) +1)^2 is not 2x +1 but 2x+1+2 root(2x). I thought I would just take every term to the secound power, such as root(2x)^2 + 1^2...

Must know property: $$(a+b)^2=a^2+2ab+b^2$$.

Check the following link for more: algebra-101576.html

Hope this helps.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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17 Apr 2014, 03:20
Bunuel wrote:
If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Diagnostic Test
Question: 16
Page: 22
Difficulty: 600

$$\sqrt{3-2x} = \sqrt{2x} +1$$

$$\sqrt{3-2x} - \sqrt{2x} = 1$$

Squaring both sides

$$3 - 2x + 2 \sqrt{(3-2x) . 2x} + 2x = 1$$

$$3 + \sqrt{24x - 16x^2} = 1$$

$$24x - 16x^2 = 4$$

$$16x^2 = 24x - 4$$

$$4x^2 = 6x - 1$$

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If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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31 Jul 2014, 04:31
Hi,

I have one question. I got to the equation 1 = 2x +\sqrt{2x} but then I squared 1- \sqrt{2x} = 2x to get to 4x^2 easily but I got stuck with the left term. So my question is how can I get the solution with this approach and additionally why did you guys square the term the other way round? It makes much more sense to me to square 2x in order to arrive at 4x^2 instead of squaring 1-2x.

I hope you understand what I'm asking.

Thanks for any help!
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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31 Jul 2014, 07:23
pipe19 wrote:
Hi,

I have one question. I got to the equation 1 = 2x +\sqrt{2x} but then I squared 1- \sqrt{2x} = 2x to get to 4x^2 easily but I got stuck with the left term. So my question is how can I get the solution with this approach and additionally why did you guys square the term the other way round? It makes much more sense to me to square 2x in order to arrive at 4x^2 instead of squaring 1-2x.

I hope you understand what I'm asking.

Thanks for any help!

Next, if you square $$1-\sqrt{2x}=2x$$ you'll get $$1 - 2\sqrt{2x}+2x=4x^2$$. As you can see we don't have this answer among the options, so we should have done the other way around.

There are two ways of solving this given above:
if-root-3-2x-root-2x-1-then-4x-135539.html#p1104029
if-root-3-2x-root-2x-1-then-4x-135539.html#p1218265
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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29 Oct 2014, 23:47
Hi guys,

I'm new to this community!

Wondering why Sqrt(3-2x)^2 got expanded to [Sqrt(3-2x)][Sqrt(3+2x)] but with (1-2x)^2 it got expanded to (1-2x)(1-2x)? I.e. the postive and negative signs. I thought whenever we get something in the form of (x-y)^2, the expansion will always be (x-y)(x+y) not (x-y)(x-y)?

Thanks a mil!
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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30 Oct 2014, 00:05
ColdSushi wrote:
Hi guys,

I'm new to this community!

Wondering why Sqrt(3-2x)^2 got expanded to [Sqrt(3-2x)][Sqrt(3+2x)] but with (1-2x)^2 it got expanded to (1-2x)(1-2x)? I.e. the postive and negative signs. I thought whenever we get something in the form of (x-y)^2, the expansion will always be (x-y)(x+y) not (x-y)(x-y)?

Thanks a mil!

$$(\sqrt{3-2x})^2 = 3-2x$$

$$(1-2x)^2 = (1-2x)(1-2x)$$

$$(x-y)^2 = x^2 - 2xy + y^2$$

$$x^2 - y^2 = (x+y)(x-y)$$

Hope this clarifies
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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30 Oct 2014, 04:02
PareshGmat wrote:
ColdSushi wrote:
Hi guys,

I'm new to this community!

Wondering why Sqrt(3-2x)^2 got expanded to [Sqrt(3-2x)][Sqrt(3+2x)] but with (1-2x)^2 it got expanded to (1-2x)(1-2x)? I.e. the postive and negative signs. I thought whenever we get something in the form of (x-y)^2, the expansion will always be (x-y)(x+y) not (x-y)(x-y)?

Thanks a mil!

$$(\sqrt{3-2x})^2 = 3-2x$$

$$(1-2x)^2 = (1-2x)(1-2x)$$

$$(x-y)^2 = x^2 - 2xy + y^2$$

$$x^2 - y^2 = (x+y)(x-y)$$

Hope this clarifies

Wow - I totally mixed up the rules - thanks for clarifying.
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If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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15 Jan 2015, 08:16
Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Hi,

Put x=1,
LHS= 1,
RHS=2.414
or LHS < RHS

Put x=0,
LHS= 1.732
RHS=1
LHS>RHS,

Thus, 0< x < 1, for equality to hold true.
or 0 < 2x < 2, squaring,
$$0 < 4x^2 < 4$$

Lets check the options now,
(A) 1 ($$0 < 4x^2 < 4$$)
(B) 4 ($$0 < 4x^2 < 4$$)
(C)2 − 2x (0< x < 1, thus, 0< 2 − 2x < 2, but $$0 < 4x^2 < 4$$)
(D) 4x − 2 (0< x < 1, thus, -2< 4x-2 < 2, but $$0 < 4x^2 < 4$$)
(E) 6x − 1 (0< x < 1, thus, -1< 6x-1 < 5, and $$0 < 4x^2 < 4$$)

Thus, only possible option which lies in the range is E.

Regards,

Hi,
Could you please explain how you got into this 0<x<1 from the LHS and RHS?
Also, the calculations of highlighted portion in the answer choice E.
TIA
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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08 Feb 2015, 06:41
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Bunuel wrote:
SOLUTION

If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

$$\sqrt{3-2x} = \sqrt{2x} +1$$ --> square both sides: $$(\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2$$ --> $$3-2x=2x+2*\sqrt{2x}+1$$ --> rearrange so that to have root at one side: $$2-4x=2*\sqrt{2x}$$ --> reduce by 2: $$1-2x=\sqrt{2x}$$ --> square again: $$(1-2x)^2=(\sqrt{2x})^2$$ --> $$1-4x+4x^2=2x$$ --> rearrange again: $$4x^2=6x-1$$.

sorry but a noob question here, i thought we aren't allowed to square both sides like this?

but instead all we can do is just to multiply both sides or divide both sides by the same number/objects etc.

cause when you square both sides, both sides is multiplying with different things right???

this part i don't get.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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04 May 2015, 12:51
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =   [#permalink] 04 May 2015, 12:51

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