If sq rt (3-2x) = sq rt (2x) + 1, then 4x^2=
A. 1
B. 4
C. 2 - 2x
D. 4x - 2
E. 6x - 1
Source: OG 12
OA after some discussion.
If sq rt (3-2x) = sq rt (2x) + 1.....
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Since none of the answer choices contains a √, we need to perform algebra that will clear all the √ symbols:chendawg wrote:If sq rt (3-2x) = sq rt (2x) + 1, then 4x^2=
A. 1
B. 4
C. 2 - 2x
D. 4x - 2
E. 6x - 1
Source: OG 12
OA after some discussion.
√(3-2x) = √2x + 1
[√(3-2x)]² = (√2x + 1)²
3 - 2x = (√2x + 1)(√2x + 1)
3 - 2x = 2x + √2x + √2x + 1
2 - 2x = 2x + 2√2x
1 - x = x + √2x
1 - 2x = √2x
(1 - 2x)² = (√2x)²
(1 - 2x)(1 - 2x) = 2x
1 - 2x - 2x + 4x² = 2x
4x² = 6x - 1.
The correct answer is E.
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- anshumishra
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√(3-2x) - √2x = 1; square both sideschendawg wrote:If sq rt (3-2x) = sq rt (2x) + 1, then 4x^2=
A. 1
B. 4
C. 2 - 2x
D. 4x - 2
E. 6x - 1
Source: OG 12
OA after some discussion.
=> 3-2x + 2x - 2√(3-2x)(2x) = 1
=> 2√(3-2x)(2x) = 2
=> √(3-2x)(2x) = 1; square both sides
=> (3-2x)*2x = 1
=> 6x - 4x^2 = 1
=> 4x^2 = 6x - 1, E
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Anshu
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OA is indeed E.
I'm not sure if I would be able to figure out to do another square under test conditions, is there a clue for future problems?
I'm not sure if I would be able to figure out to do another square under test conditions, is there a clue for future problems?
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Squaring both sides of the equation, we have:chendawg wrote:If sq rt (3-2x) = sq rt (2x) + 1, then 4x^2=
A. 1
B. 4
C. 2 - 2x
D. 4x - 2
E. 6x - 1
3 - 2x = 2x + 1 + 2√(2x)
2 - 4x = 2√(2x)
1 - 2x = √(2x)
Squaring both sides of the equation again, we have:
1 + 4x^2 - 4x = 2x
4x^2 = 6x - 1
Answer:E
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