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

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GRE Instructor
Joined: 10 Apr 2015
Posts: 2311
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Kudos [?]: 2208 [1] , given: 27

3x-1=[square_root]8x^2-4x+9[/square_root] [#permalink]  10 Aug 2019, 07:14
1
KUDOS
Expert's post
00:00

Question Stats:

66% (00:00) correct 33% (00:36) wrong based on 3 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

_________________

Brent Hanneson – Creator of greenlighttestprep.com

GRE Instructor
Joined: 10 Apr 2015
Posts: 2311
Followers: 73

Kudos [?]: 2208 [1] , given: 27

Re: 3x-1=[square_root]8x^2-4x+9[/square_root] [#permalink]  10 Aug 2019, 07:19
1
KUDOS
Expert's post
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

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com

Re: 3x-1=[square_root]8x^2-4x+9[/square_root]   [#permalink] 10 Aug 2019, 07:19
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