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Sub 505 (Easy)|   Algebra|      
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Bunuel
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If (3 - 2x)^(1/2) = 1, then what is the value of (3 – 2x) + (3 – 2x)^2 10 Apr 2019, 23:21
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90% (00:47) correct 10% (00:54) wrong based on 92 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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C
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If (3 - 2x)^(1/2) = 1, then what is the value of (3 – 2x) + (3 – 2x)^2 11 Apr 2019, 01:01
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Bunuel
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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\((3 – 2x) + (3 – 2x)^2\)
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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If (3 - 2x)^(1/2) = 1, then what is the value of (3 – 2x) + (3 – 2x)^2 11 Apr 2019, 11:20
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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


\((3 – 2x) + (3 – 2x)^2\)
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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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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If (3 - 2x)^(1/2) = 1, then what is the value of (3 – 2x) + (3 – 2x)^2 11 Apr 2019, 11:33
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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
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Bunuel
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
Show more
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If (3 - 2x)^(1/2) = 1, then what is the value of (3 – 2x) + (3 – 2x)^2 11 Apr 2019, 11:42
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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)

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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 16 Apr 2019, 09:54
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thealpine


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?
Show more
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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