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

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Joined: 02 Sep 2009
Posts: 55272
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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30 May 2015, 04:43
reto 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

Diagnostic Test
Question: 16
Page: 22
Difficulty: 600

Why is this question categorized as difficulty 700 above and 600 in the question steam? What's the actual difficulty?

If you square the expression $$\sqrt{2x} +1$$ and there are NO paranthesis, why is it not correct to square each term seperately to get 2x + 1?

Thank you so much

1. 600 difficulty was given when posted. 700 difficulty is based on users' answers on the question.
2. (a + b)^2 generally does not equal to a^2 + b^2. I'ts pretty basic actually.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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30 May 2015, 05:04
Bunuel wrote:
reto 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

Diagnostic Test
Question: 16
Page: 22
Difficulty: 600

Why is this question categorized as difficulty 700 above and 600 in the question steam? What's the actual difficulty?

If you square the expression $$\sqrt{2x} +1$$ and there are NO paranthesis, why is it not correct to square each term seperately to get 2x + 1?

Thank you so much

1. 600 difficulty was given when posted. 700 difficulty is based on users' answers on the question.
2. (a + b)^2 generally does not equal to a^2 + b^2. I'ts pretty basic actually.

Thanks. Thats what I meant in the question there are no paranthesis, but when you squared it in your solution, you gave the expression extra paranthesis - why? I know it's not the same. But I don't understand why you completed it when squaring...
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Posts: 55272
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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30 May 2015, 05:21
1
reto wrote:
Thanks. Thats what I meant in the question there are no paranthesis, but when you squared it in your solution, you gave the expression extra paranthesis - why? I know it's not the same. But I don't understand why you completed it when squaring...

Squaring means multiplying by the same number, so when you square a + b, you get (a + b)(a + b) = (a + b)^2.

I'd advice to brush up fundamentals before practicing questions, especially, hard ones.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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20 Mar 2016, 18:25
1
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)-1 = sqrt(2x) | square everything
2x = 3-2x - 2*sqrt(3-2x)+1
2x=4 - 2x - 2*sqrt(3-2x)
4-4x=2*sqrt(3-2x) | square everything again
16+16x^2 -32x = 12 - 8x | add 8x to both sides, subtract 12 from both sides
16x^2 - 24x +4 = 0 | divide by 4
4x^2 -6x +1 = 0 | add 6x to both sides and subtract 1:
4x^2 = 6x-1

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

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08 Mar 2018, 18:48
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

Main Idea:While simplifying keep the more complex root expression alone on one side and simplify.

Details: sqrt(3-2x)=sqrt(2) +1.

Squaring on both sides: 3-2x=2x+1+2*sqrt(2x)

Simplifying we get (4x)^2 = 6x-1

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

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02 May 2018, 08:51
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: $$4x^2=6x-1$$.

how from this $$(\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2$$ we get this $$3-2x=2x+2*\sqrt{2x}+1$$ ?

thank you for your kind attention to my question
Math Expert
Joined: 02 Sep 2009
Posts: 55272
If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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02 May 2018, 09:17
dave13 wrote:
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: $$4x^2=6x-1$$.

how from this $$(\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2$$ we get this $$3-2x=2x+2*\sqrt{2x}+1$$ ?

thank you for your kind attention to my question

1. What does $$(\sqrt{a})^2$$ equal to?
2. What does $$(a + b)^2$$ equal to?

If you can answer these questions, you should be able to figure out your doubt.
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If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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02 May 2018, 10:36
Bunuel wrote:
dave13 wrote:
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: $$4x^2=6x-1$$.

how from this $$(\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2$$ we get this $$3-2x=2x+2*\sqrt{2x}+1$$ ?

thank you for your kind attention to my question

1. What does $$(\sqrt{a})^2$$ equal to?
2. What does $$(a + b)^2$$ equal to?

If you can answer these questions, you should be able to figure out your doubt.

Hi Bunuel

1. What does $$(\sqrt{a})^2$$ equal to? it equals $$a$$

2. What does $$(a + b)^2$$ equal to? it equals this $$(a+b)(a+b) = a^2+ab+ba+b^2 = a^2+2ab+b^2$$

now let me get back to the question

$$(\sqrt{3-2x})^2 = (\sqrt{2x} +1)^2$$ square both sides (using this formula $$(\sqrt{a})^2$$ = $$a$$

so we get $$(3-2x)^2 = (2x-1)^2$$

now using this formula $$(a + b)^2$$ = $$a^2+2ab+b^2$$

we get $$(3-2x) ( 3-2x) = (2x+1) (2x+1)$$ so from here we get $$10=16x$$:?

can you please explain what am i doing wrong knowing formula is one thing, but applying it correctly is quite challenging for me ......sometimes
Math Expert
Joined: 02 Sep 2009
Posts: 55272
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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02 May 2018, 12:28
1
dave13 wrote:
Bunuel wrote:
dave13 wrote:
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: $$4x^2=6x-1$$.

how from this $$(\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2$$ we get this $$3-2x=2x+2*\sqrt{2x}+1$$ ?

thank you for your kind attention to my question

1. What does $$(\sqrt{a})^2$$ equal to?
2. What does $$(a + b)^2$$ equal to?

If you can answer these questions, you should be able to figure out your doubt.

Hi Bunuel

1. What does $$(\sqrt{a})^2$$ equal to? it equals $$a$$

2. What does $$(a + b)^2$$ equal to? it equals this $$(a+b)(a+b) = a^2+ab+ba+b^2 = a^2+2ab+b^2$$

now let me get back to the question

$$(\sqrt{3-2x})^2 = (\sqrt{2x} +1)^2$$ square both sides (using this formula $$(\sqrt{a})^2$$ = $$a$$

so we get $$(3-2x)^2 = (2x-1)^2$$

now using this formula $$(a + b)^2$$ = $$a^2+2ab+b^2$$

we get $$(3-2x) ( 3-2x) = (2x+1) (2x+1)$$ so from here we get $$10=16x$$:?

can you please explain what am i doing wrong knowing formula is one thing, but applying it correctly is quite challenging for me ......sometimes

1. Yes, $$(\sqrt{a})^2=a$$, hence $$(\sqrt{3-2x})^2 =3-2x$$.

2. Here you are also correct: $$(a + b)^2=a^2+2ab+b^2$$, hence $$(\sqrt{2x} +1)^2=(\sqrt{2x})^2 + 2*\sqrt{2x}*1+1^2=2x+2*\sqrt{2x}+1$$.

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

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03 May 2018, 10:31
Bunuel, it makes perfect sense !

i applied this formula $$(a - b)^2$$ = $$a^2-2ab+b^2$$ to $$(1-2x)^2=(\sqrt{2x})^2$$

so i get this $$1^2 - 2 * 1 *(-2x) +2x^2 = 2x$$ ----> $$1 -2-2x+4x=2x$$ --> $$1 -2+2x= 2x$$ ---> $$-1 =0$$:? what am i doing wrong , can you please explain in two or three words

many thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 55272
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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03 May 2018, 11:16
1
dave13 wrote:
Bunuel, it makes perfect sense !

i applied this formula $$(a - b)^2$$ = $$a^2-2ab+b^2$$ to $$(1-2x)^2=(\sqrt{2x})^2$$

so i get this 1^2 - 2 * 1 *(-2x) +2x^2 = 2x ----> $$1 -2-2x+4x=2x$$ --> $$1 -2+2x= 2x$$ ---> $$-1 =0$$:? what am i doing wrong , can you please explain in two or three words

many thanks!

I highlighted the mistakes.

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

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03 May 2018, 12:19
Bunuel wrote:
dave13 wrote:
Bunuel, it makes perfect sense !

i applied this formula $$(a - b)^2$$ = $$a^2-2ab+b^2$$ to $$(1-2x)^2=(\sqrt{2x})^2$$

so i get this 1^2 - 2 * 1 *(-2x) +2x^2 = 2x ----> $$1 -2-2x+4x=2x$$ --> $$1 -2+2x= 2x$$ ---> $$-1 =0$$:? what am i doing wrong , can you please explain in two or three words

many thanks!

I highlighted the mistakes.

$$(1 - 2x)^2 = 1 - 2*2x + (2x)^2 = 1 - 4x + 4x^2$$

Bunuel thank you ! just last question

how did you convert this $$(2x)^2$$ and got $$4x^2$$

$$(2x)^2$$ equals $$4x$$ and not $$4x^2$$

please explain this part and I will be more than happy
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Posts: 55272
Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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03 May 2018, 12:21
1
dave13 wrote:
Bunuel wrote:
dave13 wrote:
Bunuel, it makes perfect sense !

i applied this formula $$(a - b)^2$$ = $$a^2-2ab+b^2$$ to $$(1-2x)^2=(\sqrt{2x})^2$$

so i get this 1^2 - 2 * 1 *(-2x) +2x^2 = 2x ----> $$1 -2-2x+4x=2x$$ --> $$1 -2+2x= 2x$$ ---> $$-1 =0$$:? what am i doing wrong , can you please explain in two or three words

many thanks!

I highlighted the mistakes.

$$(1 - 2x)^2 = 1 - 2*2x + (2x)^2 = 1 - 4x + 4x^2$$

Bunuel thank you ! just last question

how did you convert this $$(2x)^2$$ and got $$4x^2$$

$$(2x)^2$$ equals $$4x$$and not $$4x^2$$

please explain this part and I will be more than happy

We have $$(2x)^2$$. Notice that we are squaring 2x, so you should square both 2 and x: $$(2x)^2=2x*2x=4x^2$$
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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14 Feb 2019, 09:58
this is tricky
the key here is to arrange the equation so that we have root at one side. when this happen, we can square and after squaring, we can have simple result.
we square the first time immediately because the left side is a root already
after getting the result fo the first squaring, we arrange so that we have root at one side.

look at the way Bunnuel do. I just explain more.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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14 Feb 2019, 10:16
1
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

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

Square once to get
3-2x = 2x + 1 + 2$$\sqrt{2x}$$
1-2x = $$\sqrt{2x}$$

Square again to get
$$(1-2x)^2$$ = 2x
4x^2 = 6x - 1

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

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18 Mar 2019, 10:07
Top Contributor
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

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

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

Expand and simplify to get: $$3-2x=2x+2\sqrt{2x}+1$$

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

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

Cheers,
Brent
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =   [#permalink] 18 Mar 2019, 10:07

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