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
_________________
Brent Hanneson – Creator of greenlighttestprep.com
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