Equation I: Quadratic Check

To determine if Equation I is quadratic, simplify it:

• Given: $x^2 + 4x + 23 = x + \frac{33}{28}$
• Rearrange terms to one side: $x^2 + 4x - x + 23 - \frac{33}{28} = 0$
• Combine like terms: $x^2 + 3x + \left( \frac{23 \times 28}{28} - \frac{33}{28} \right) = 0$
• Simplify: $x^2 + 3x + \frac{644 - 33}{28} = 0$
• Final form: $x^2 + 3x + \frac{611}{28} = 0$

This is a quadratic equation as the highest power of $x$ is 2 ($a=1 \neq 0$).

Equation II: Quadratic Check

Simplify Equation II:

• Given: $\frac{2}{3}x^2 + 6x + 2 = x^2 + 8$
• Rearrange terms: $\frac{2}{3}x^2 - x^2 + 6x + 2 - 8 = 0$
• Combine like terms: $\left( \frac{2}{3} - 1 \right)x^2 + 6x - 6 = 0$
• Simplify: $-\frac{1}{3}x^2 + 6x - 6 = 0$

This is a quadratic equation because the highest power of $x$ is 2 ($a=-\frac{1}{3} \neq 0$).

Equation III: Quadratic Check

Simplify Equation III:

• Given: $6x^2 + 4x + 18 = 6x^2 - 6x$
• Rearrange terms: $6x^2 - 6x^2 + 4x + 6x + 18 = 0$
• Combine like terms: $(6-6)x^2 + (4+6)x + 18 = 0$
• Simplify: $0x^2 + 10x + 18 = 0$, which becomes $10x + 18 = 0$

This is a linear equation, not quadratic, as the $x^2$ terms cancel out ($a=0$).

Conclusion: Quadratic Equations

Equations I and II simplify to forms where the highest power of $x$ is 2, making them quadratic equations. Equation III simplifies to a linear equation.

Therefore, the quadratic equations are I & II.