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I'm sorry, I just realised I put this in the wrong subsection. While I figure out how to fix that, please have a look anyway.

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Given x [itex]\inℝ[/itex]

And s =[itex]\frac{4(x^{2}) + 3}{2x-1}[/itex]

Prove that [itex]s^{2}[/itex] -4s - 12 ≥ 0

The discriminant Δ, (in order for which to be real must be ≥ 0)

b^2 - 4ac ≥ 0

Doing the algebra isn't the problem, I'm having trouble understanding the question itself. For this sort of proof, don't I need to work with

s =[itex]\frac{4(x^{2}) + 3}{2x-1}[/itex]

instead of the statement to be proven, which is [itex]s^{2}[/itex] -4s - 12 ≥ 0?

In which case, how do I apply the b^2 - 4ac rule with the linear equation part in the denominator?

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**1. Homework Statement**Given x [itex]\inℝ[/itex]

And s =[itex]\frac{4(x^{2}) + 3}{2x-1}[/itex]

Prove that [itex]s^{2}[/itex] -4s - 12 ≥ 0

**2. Homework Equations**The discriminant Δ, (in order for which to be real must be ≥ 0)

b^2 - 4ac ≥ 0

**3. The Attempt at a Solution**Doing the algebra isn't the problem, I'm having trouble understanding the question itself. For this sort of proof, don't I need to work with

s =[itex]\frac{4(x^{2}) + 3}{2x-1}[/itex]

instead of the statement to be proven, which is [itex]s^{2}[/itex] -4s - 12 ≥ 0?

In which case, how do I apply the b^2 - 4ac rule with the linear equation part in the denominator?

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