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**Solutions:**
- **a)** \( x = -3 \) and \( x = -4 \)
- **b)** \( x = 1 \) and \( x = -1 \)
Solution
Sure, let's solve each equation step by step.
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### **a) \( x^{2} + 7x + 12 = 0 \)**
This is a quadratic equation of the form \( ax^{2} + bx + c = 0 \). To solve it, we can factor the quadratic expression.
1. **Factor the quadratic:**
\( x^{2} + 7x + 12 \)
We need two numbers that multiply to **12** and add up to **7**. Those numbers are **3** and **4**.
So, we can rewrite the quadratic as:
\( (x + 3)(x + 4) = 0 \)
2. **Find the roots:**
Set each factor equal to zero:
- \( x + 3 = 0 \) ⇒ \( x = -3 \)
- \( x + 4 = 0 \) ⇒ \( x = -4 \)
**Solutions:** \( x = -3 \) and \( x = -4 \)
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### **b) \( x^{2} - 1 = 1 - x^{2} \)**
Let's solve for \( x \):
1. **Bring all terms to one side to set the equation to zero:**
\( x^{2} - 1 - 1 + x^{2} = 0 \)
Simplify:
\( 2x^{2} - 2 = 0 \)
2. **Solve for \( x^{2} \):**
\( 2x^{2} = 2 \)
Divide both sides by 2:
\( x^{2} = 1 \)
3. **Take the square root of both sides:**
\( x = \pm \sqrt{1} \)
\( x = \pm 1 \)
**Solutions:** \( x = 1 \) and \( x = -1 \)
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**Summary of Solutions:**
- **a)** \( x = -3 \) and \( x = -4 \)
- **b)** \( x = 1 \) and \( x = -1 \)
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Bonus Knowledge
Let's unravel the mathematical mysteries behind these equations! To solve the first equation \( x^{2}+7x+12=0 \), you can factor it into \( (x+3)(x+4)=0 \). This gives you the solutions \( x=-3 \) and \( x=-4 \). For the second equation \( x^{2}-1=1-x^{2} \), start by rearranging it to combine like terms: \( x^{2} + x^{2} - 1 - 1 = 0 \), which simplifies to \( 2x^{2} - 2 = 0 \). Dividing through by 2 leads to \( x^{2} = 1 \), giving you the solutions \( x=1 \) and \( x=-1 \).
