Let's solve each quadratic equation step-by-step.
1. \((x+3)(x−1)=0\)
To find the solutions, we set each factor to zero:
- \(x + 3 = 0 \Rightarrow x = -3\)
- \(x - 1 = 0 \Rightarrow x = 1\)
Answer: x = -3 and x = 1
2. \((15x−5)(7+x)=0\)
Setting each factor to zero:
- \(15x - 5 = 0 \Rightarrow 15x = 5 \Rightarrow x = \frac{1}{3}\)
- \(7 + x = 0 \Rightarrow x = -7\)
Answer: x = -7 and x = \(\frac{1}{3}\)
3. \((13x−2)(x−34)=0\)
Setting each factor to zero:
- \(13x - 2 = 0 \Rightarrow 13x = 2 \Rightarrow x = \frac{2}{13}\)
- \(x - 34 = 0 \Rightarrow x = 34\)
Answer: x = \(\frac{2}{13}\) and x = 34
4. \(30=x^2+13x\)
First, rearrange the equation into standard form: \[ x^2 + 13x - 30 = 0 \]
Now, we can factor this quadratic. We are looking for two numbers that multiply to \(-30\) and add up to \(13\). These numbers are \(15\) and \(-2\). Factoring: \[(x + 15)(x - 2) = 0\]
Setting each factor to zero:
- \(x + 15 = 0 \Rightarrow x = -15\)
- \(x - 2 = 0 \Rightarrow x = 2\)
Answer: x = -15 and x = 2
Summary of Solutions:
- \(x = -3\) and \(x = 1\)
- \(x = -7\) and \(x = \frac{1}{3}\)
- \(x = \frac{2}{13}\) and \(x = 34\)
- \(x = -15\) and \(x = 2\)