Let's solve each quadratic equation step by step.
Problem 1: \( (x + 3)(x - 1) = 0 \)
To solve this equation, we set each factor to zero:
- \( x + 3 = 0 \) → \( x = -3 \)
- \( x - 1 = 0 \) → \( x = 1 \)
Thus, the answers are:
- \( x = -3 \)
- \( x = 1 \)
Problem 2: \( (15x - 5)(7 + x) = 0 \)
Set each factor to zero:
-
\( 15x - 5 = 0 \)
- \( 15x = 5 \)
- \( x = \frac{5}{15} = \frac{1}{3} \)
-
\( 7 + x = 0 \)
- \( x = -7 \)
Thus, the answers are:
- \( x = -7 \)
- \( x = \frac{1}{3} \)
Problem 3: \( (13x - 2)(x - 34) = 0 \)
Set each factor to zero:
-
\( 13x - 2 = 0 \)
- \( 13x = 2 \)
- \( x = \frac{2}{13} \)
-
\( x - 34 = 0 \)
- \( x = 34 \)
Thus, the answers are:
- \( x = \frac{2}{13} \)
- \( x = 34 \)
Problem 4: \( 30 = x^2 + 13x \)
Rearrange to standard form:
\[ x^2 + 13x - 30 = 0 \]
Now we can factor or use the quadratic formula. Let's factor this:
We need two numbers that multiply to \(-30\) and add up to \(13\):
- The numbers are \(15\) and \(-2\).
So we can factor the equation:
\[ (x + 15)(x - 2) = 0 \]
Set each factor to zero:
-
\( x + 15 = 0 \)
- \( x = -15 \)
-
\( x - 2 = 0 \)
- \( x = 2 \)
Thus, the answers are:
- \( x = -15 \)
- \( x = 2 \)
Summary of Solutions:
-
\( (x + 3)(x - 1) = 0 \)
\( x = -3 \) and \( x = 1 \) -
\( (15x - 5)(7 + x) = 0 \)
\( x = -7 \) and \( x = \frac{1}{3} \) -
\( (13x - 2)(x - 34) = 0 \)
\( x = \frac{2}{13} \) and \( x = 34 \) -
\( 30 = x^2 + 13x \)
\( x = -15 \) and \( x = 2 \)