Determine whether the solutions to the following equations satisfies 𝑥>0, 𝑥=0 or 𝑥>0.𝑥<0𝑥=0𝑥>0 4𝑥−12=𝑥2+2−2𝑥+3=10+4𝑥2𝑥+7=7−5𝑥 5𝑥+23=𝑥−12

To determine whether the solutions to the provided equations satisfy the specified inequalities $x > 0$, $x = 0$, or $x < 0$, we will follow the steps outlined below for each equation.

1. Break Down the Problem

We have the following equations to solve:

1. $4x - 12 = 0$
2. $x^2 + 2 - 2x + 3 = 10 + 4x$
3. $x + 7 = 7 - 5x$
4. $5x + 23 = x - 12$

2. Relevant Concepts

We will solve each equation step-by-step to find the value of $x$ and then determine if it satisfies $x > 0$, $x = 0$, or $x < 0$.

3. Analysis and Detail

3.1 Solve Equation 1: $4x - 12 = 0$

• Step 1: Add 12 to both sides. $4x = 12$
• Step 2: Divide by 4. $x = 3$
• Check: $3 > 0$ (satisfies $x > 0$).

3.2 Solve Equation 2: $x^2 + 2 - 2x + 3 = 10 + 4x$

• Step 1: Simplify the left side: $x^2 - 2x + 5 = 10 + 4x$

• Step 2: Move all terms to one side: $x^2 - 6x - 5 = 0$

• Step 3: Factor or use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

  • Here, $a = 1, b = -6, c = -5$: $x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-5)}}{2(1)} = \frac{6 \pm \sqrt{36 + 20}}{2} = \frac{6 \pm \sqrt{56}}{2} = \frac{6 \pm 2\sqrt{14}}{2} = 3 \pm \sqrt{14}$

• The solutions: $x = 3 + \sqrt{14}$ and $x = 3 - \sqrt{14}$.

• Check:

  • $3 + \sqrt{14} > 0$ (satisfies $x > 0$).
  • $3 - \sqrt{14} < 0$ (satisfies $x < 0$).

3.3 Solve Equation 3: $x + 7 = 7 - 5x$

• Step 1: Move $x$ to the right side: $7 + 5x = 7 - x$
• Step 2: Combine like terms: $6x = 0$
• Step 3: Divide by 6: $x = 0$
• Check: $x = 0$ (satisfies $x = 0$).

3.4 Solve Equation 4: $5x + 23 = x - 12$

• Step 1: Move $x$ to the left side: $5x - x = -12 - 23$
• Step 2: Combine like terms: $4x = -35$
• Step 3: Divide by 4: $x = -\frac{35}{4}$
• Check: $-\frac{35}{4} < 0$ (satisfies $x < 0$).

4. Verify and Summarize

The solutions and their respective checks yield:

• From Equation 1: $x = 3$ satisfies $x > 0$.
• From Equation 2: $x = 3 + \sqrt{14}$ satisfies $x > 0$ and $x = 3 - \sqrt{14}$ satisfies $x < 0$.
• From Equation 3: $x = 0$ satisfies $x = 0$.
• From Equation 4: $x = -\frac{35}{4}$ satisfies $x < 0$.

Final Answer

The solutions satisfy the conditions as follows:

1. $x = 3: \, x > 0$
2. $x = 3 + \sqrt{14}: \, x > 0$
3. $x = 3 - \sqrt{14}: \, x < 0$
4. $x = 0: \, x = 0$
5. $x = -\frac{35}{4}: \, x < 0$