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# If (3 - 2x)^(1/2) = 1, then what is the value of (3 – 2x) + (3 – 2x)^2

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Joined: 02 Sep 2009
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If (3 - 2x)^(1/2) = 1, then what is the value of (3 – 2x) + (3 – 2x)^2  [#permalink]

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10 Apr 2019, 23:21
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5% (low)

Question Stats:

94% (00:45) correct 6% (00:22) wrong based on 31 sessions

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If $$\sqrt{3 - 2x} = 1$$, then what is the value of $$(3 – 2x) + (3 – 2x)^2$$ ?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

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Re: If (3 - 2x)^(1/2) = 1, then what is the value of (3 – 2x) + (3 – 2x)^2  [#permalink]

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11 Apr 2019, 01:01
Bunuel wrote:
If $$\sqrt{3 - 2x} = 1$$, then what is the value of $$(3 – 2x) + (3 – 2x)^2$$ ?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

$$(3 – 2x) + (3 – 2x)^2$$
3-2x= $$\sqrt{3 - 2x} * [m]\sqrt{3 - 2x} so [m](3 – 2x) + (3 – 2x)^2$$ = $$\sqrt{3 - 2x} * [m]\sqrt{3 - 2x} + ( [m]\sqrt{3 - 2x} * [m]\sqrt{3 - 2x} )^2 given [m]\sqrt{3 - 2x} = 1$$
so 1+1 = 2
IMO C
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Joined: 05 Apr 2019
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If (3 - 2x)^(1/2) = 1, then what is the value of (3 – 2x) + (3 – 2x)^2  [#permalink]

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11 Apr 2019, 11:20
Archit3110 wrote:
Bunuel wrote:
If $$\sqrt{3 - 2x} = 1$$, then what is the value of $$(3 – 2x) + (3 – 2x)^2$$ ?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

$$(3 – 2x) + (3 – 2x)^2$$
3-2x= $$\sqrt{3 - 2x} * [m]\sqrt{3 - 2x} so [m](3 – 2x) + (3 – 2x)^2$$ = $$\sqrt{3 - 2x} * [m]\sqrt{3 - 2x} + ( [m]\sqrt{3 - 2x} * [m]\sqrt{3 - 2x} )^2 given [m]\sqrt{3 - 2x} = 1$$
so 1+1 = 2
IMO C

Why not square both sides of the first equation to eliminate the square root? and then solve.

I got 0 as my answer
CEO
Joined: 18 Aug 2017
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Location: India
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Re: If (3 - 2x)^(1/2) = 1, then what is the value of (3 – 2x) + (3 – 2x)^2  [#permalink]

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11 Apr 2019, 11:33
thealpine ; squaring both side will give +/- values and this question we can solve without squaring as well √x*√x=x so ; 3-2x = (√3-2x) * (√3-2x)

thealpine wrote:
Archit3110 wrote:
Bunuel wrote:
If $$\sqrt{3 - 2x} = 1$$, then what is the value of $$(3 – 2x) + (3 – 2x)^2$$ ?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

$$(3 – 2x) + (3 – 2x)^2$$
3-2x= $$\sqrt{3 - 2x} * [m]\sqrt{3 - 2x} so [m](3 – 2x) + (3 – 2x)^2$$ = $$\sqrt{3 - 2x} * [m]\sqrt{3 - 2x} + ( [m]\sqrt{3 - 2x} * [m]\sqrt{3 - 2x} )^2 given [m]\sqrt{3 - 2x} = 1$$
so 1+1 = 2
IMO C

Why not square both sides of the first equation to eliminate the square root? and then solve.

I got 0 as my answer

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Intern
Joined: 05 Apr 2019
Posts: 7
Re: If (3 - 2x)^(1/2) = 1, then what is the value of (3 – 2x) + (3 – 2x)^2  [#permalink]

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11 Apr 2019, 11:42
Archit3110 wrote:
thealpine ; squaring both side will give +/- values and this question we can solve without squaring as well √x*√x=x so ; 3-2x = (√3-2x) * (√3-2x)

I heard this on a teaching platform, that whenever the GMAT gives your the square root on the prompt, you can/it is okay to only consider positive values. However, if the square root is something you come across while solving for a problem then you have to consider both positive and negative values.

Heard this from MikeMcGarry

Thoughts on this?
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If (3 - 2x)^(1/2) = 1, then what is the value of (3 – 2x) + (3 – 2x)^2  [#permalink]

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16 Apr 2019, 09:54
thealpine wrote:

I heard this on a teaching platform, that whenever the GMAT gives your the square root on the prompt, you can/it is okay to only consider positive values. However, if the square root is something you come across while solving for a problem then you have to consider both positive and negative values.

Heard this from MikeMcGarry

Thoughts on this?

Indeed, it is correct to consider only positive values for the square roots on the prompt on GMAT; otherwise, you would be handling with complex roots.

It is important to bear in mind that while a "regular square root" and a "square root of some variable" may seem to be the same case, there is a difference.
Numbers have a real and an imaginary part - which is often ignored by us - but in this specific case the imaginary part plays an important role.

While the "regular square root" has an $$i^0$$ form, the variable solving find positive numbers from both $$i^0$$ and a combination of $$i^2$$ forms - which will multiply as a $$i^4$$ -, as the imaginary part follows the following power pattern:

$$i^0 = 1$$
$$i^1 = i$$
$$i^2 = – 1$$
$$i^3 = i^2 * i = (–1) * i = –i$$
$$i^4 = i^2 * i^2 = (–1) * (– 1) = 1$$
...

Having that in mind, as you may have realized, we can be sure that $$√1 = 1$$, since it is in the $$√(1*i^0)$$ form. On the hand, there is no certain when it comes to $$x^2=1$$ as it could be both:
$$(1*i^0)^2 = (1*1)^2 = 1^2 = 1$$
or
$$(1*i^2)^2 = (1*-1)^2 = (-1)^2 = 1$$

Hope it helps!

As for the problem, I would just consider that 1 to any power equals 1. Hence, the given expression would be the same of $$√1$$ and we could simply do
$$(3–2x)+(3–2x)^2 = 1 + 1^2 = 2$$
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If (3 - 2x)^(1/2) = 1, then what is the value of (3 – 2x) + (3 – 2x)^2   [#permalink] 16 Apr 2019, 09:54
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