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705-805 (Hard)|   Algebra|   Roots|                        
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Bunuel
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 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
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E
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 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: \(4x^2=6x-1\).

Answer: E.
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = Updated on: 20 Jul 2020, 11:01
Originally posted by BrentGMATPrepNow on 18 Mar 2019, 10:07.
Last edited by BrentGMATPrepNow on 20 Jul 2020, 11:01, edited 2 times in total.
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Bunuel
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

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

GIVEN: \(\sqrt{3-2x} = \sqrt{2x} +1\)

Square both sides to get: \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\)

In other words: \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)(\sqrt{2x} +1)\)

Use FOIL to expand right side: \((\sqrt{3-2x})^2 =(\sqrt{2x})^2 + 1\sqrt{2x} + 1\sqrt{2x} + 1^2\)

Simplify right side to get: \(3-2x=2x+2\sqrt{2x}+1\)

Subtract 1 from both sides to get: \(2-2x=2x+2\sqrt{2x}\)

Subtract 2x from both sides to get: \(2-4x=2\sqrt{2x}\)

Divide both sides by 2 to get: \(1-2x=\sqrt{2x}\)

Square both sides to get: \((1-2x)^2=(\sqrt{2x})^2\)

Expand and simplify to get: \(1-4x+4x^2=2x\)

Add 4x to both sides: \(1 + 4x^2=6x\)

Subtract 1 from both sides to get: \(4x^2=6x-1\)

Answer: E

Cheers,
Brent
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 09 Jul 2012, 06:16
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Bunuel
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

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

Answer (E),

Regards,
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 09 Jul 2012, 21:48
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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 \(4x^2=6x-1\)

Answer (E).

Regards,

PS: But that isn't much fun. :twisted:
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 29 Apr 2013, 06:40
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Bunuel
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

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

Show more

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

Please help
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 29 Apr 2013, 06:56
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Bunuel
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

Please help
Show more

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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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 31 Jul 2014, 04:31
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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. :wink:

Thanks for any help!
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pipe19
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. :wink:

Thanks for any help!
Show more

First of all please read this:

rules-for-posting-please-read-this-before-posting-133935.html#p1096628 (Writing Mathematical Formulas on the Forum)

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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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 08 Feb 2015, 06:41
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Bunuel
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: \(4x^2=6x-1\).

Answer: E.
Show more

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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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 05 May 2015, 07:19
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Bunuel
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: \(4x^2=6x-1\).

Answer: E.

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.
Show more

Hi melaos,

You are right when you say that we need to multiply or divide by the same number/object on both sides of an equation. Let me tell you why squaring both sides of an equation is a perfectly acceptable way.

Consider an equation a = b. This equation tells us that the value of a is the same as value of b. When we square a number, we multiply the number by itself. So, squaring this equation would look like a*a = b* b. Now, you would be tempted to say that we are multiplying different numbers on both sides of the equation.

Before you say so, look back at our original equation which says a = b. Hence, when we multiply a on LHS and b on RHS we are multiplying both sides by variables which have the same values.
Hence, squaring both sides of an equation is allowed.

Hope its clear!

Regards
Harsh
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 14 Jan 2023, 02:43
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Asked: If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

\(3 - 2x = 2x + 1 + 2\sqrt{2x}\)
\(2 - 4x = 2\sqrt{2x}\)
\((1-2x)^2 = 4x^2 + 1 - 4x = 2x\)
\(4x^2 = 6x - 1\)

IMO E
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 02 Dec 2023, 01:55
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Bunuel.. Why u r not proceeding here with modulus while solving square root.
As square of (3-2x)^(1/2) should be written as mod(3-2x)..??

Please heldp
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 02 Dec 2023, 02:04
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amitarya
Bunuel.. Why u r not proceeding here with modulus while solving square root.
As square of (3-2x)^(1/2) should be written as mod(3-2x)..??

Please heldp
Show more

\(\sqrt{a^2} = |a|\). Do we have the square of an expression in \(\sqrt{3-2x}\)? No. Hence, \(\sqrt{3-2x}\) does not equal to \(|3-2x|\). If it were \(\sqrt{(3-2x)^2}\), then it would be equal to \(|3-2x|\).
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 02 Dec 2023, 02:18
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Bunuel
amitarya
Bunuel.. Why u r not proceeding here with modulus while solving square root.
As square of (3-2x)^(1/2) should be written as mod(3-2x)..??

Please heldp

\(\sqrt{a^2} = |a|\). Do we have the square of an expression in \(\sqrt{3-2x}\)? No. Hence, \(\sqrt{3-2x}\) does not equal to \(|3-2x|\). If it were \(\sqrt{(3-2x)^2}\), then it would be equal to \(|3-2x|\).
Show more


Bunuel.. Got it.. Thankyou so much for clarification..
You are always a saviour.. :)
:)
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 21 Mar 2024, 23:47
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3-2x=2x+1+2rt(2x)
2-4x=2rt(2x)
4+16x^2-16x=8x
16x^2=24x-4
4x^2=6x-1
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 29 May 2024, 20:36
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­Straighforward algebra stuff we love to do:

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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 30 Aug 2025, 17:05
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Sqrt 3-2x = sqrt 2x +1
3-2x=2x+2sqrt 2x +1
0=4x+2sqrt 2x -2
0=2x + sqrt 2x -1
-2x+1 = sqrt 2x
4x^2-4x+1=2x
4x^2=6x-1

E

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

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1
Show more
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 09 Dec 2025, 08:01
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cyberjadugar

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.

Answer (E),

Regards,
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Can you please explain how you eliminated opt C and D?
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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 = 25 Mar 2026, 11:17
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Bunuel
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1
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Thinking should be to reach goal \( 4x^2 \) best way to do it is segregate radicals and square.

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

Squaring on both sides
\((\sqrt{3-2x})^2 = (\sqrt{2x} +1)^2\)

\( 3 -2x = 2\sqrt{2x} + 1 + 2x\)
\( 2 -4x = 2\sqrt{2x} \)
\( 1 -2x = \sqrt{2x} \)

Squaring on both sides

\( 1 - 4x + 4x^2 = 2x \)

\( 4x^2 = 6x - 1\)

Correct Answer : E
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