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