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3x-1=[square_root]8x^2-4x+9[/square_root]

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3x-1=[square_root]8x^2-4x+9[/square_root] [#permalink] New post 10 Aug 2019, 07:14
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Question Stats:

30% (00:41) correct 69% (01:24) wrong based on 13 sessions
\(3x-1=\sqrt{8x^2-4x+9}\)

Quantity A
Quantity B
\(x\)
\(2\)


A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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1 KUDOS received
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Re: 3x-1=[square_root]8x^2-4x+9[/square_root] [#permalink] New post 10 Aug 2019, 07:19
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GreenlightTestPrep wrote:
\(3x-1=\sqrt{8x^2-4x+9}\)

Quantity A
Quantity B
x
2


A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


Given: \(3x-1=\sqrt{8x^2-4x+9}\)

Square both sides: \((3x-1)^2=(\sqrt{8x^2-4x+9})^2\)

Expand left side and simplify right side: \(9x^2-6x+1=8x^2-4x+9\)

Subtract \(8x^2\) from both sides: \(x^2-6x+1=-4x+9\)

Add \(4x\) to both sides: \(x^2-2x+1=9\)

Subtract \(9\) from both sides: \(x^2-2x-8=0\)

Factor: \((x-4)(x+2)=0\)

Solve: \(x=4\) or \(x=-2\)

KEY STEP: Check for extraneous roots by testing both possible solutions.

First, plug \(x=4\) into original equation to get: \(3(4)-1=\sqrt{8(4^2)-4(4)+9}\)
Simplify: \(11=\sqrt{121}\)
Works!
So, \(x=4\) is a valid solution


Now plug \(x=-2\) into original equation to get: \(3(-2)-1=\sqrt{8(-2)^2-4(-2)+9}\)
Simplify: \(-7=\sqrt{49}\)
No good! \(\sqrt{49}=7\), not \(-7\)
So, \(x=-2\) is an extraneous root.

Since there's only one valid solution, we get:
Quantity A: 4
Quantity B: 2

Answer: A

Cheers,
Brent
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Re: 3x-1=[square_root]8x^2-4x+9[/square_root] [#permalink] New post 06 Sep 2019, 19:07
GreenlightTestPrep wrote:
\(3x-1=\sqrt{8x^2-4x+9}\)

Quantity A
Quantity B
\(x\)
\(2\)


A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

After solving and simplifying the above-mentioned equation, we get the roots as -2 and 4.
Here, -2 is not applicable as it is not matching up with the given equation. Only 4 is matching up, which is greater than the other option.
Hence option A.
Re: 3x-1=[square_root]8x^2-4x+9[/square_root]   [#permalink] 06 Sep 2019, 19:07
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