Question

Solve each quadratic equation. Write your answer as a simplified fraction when applicable. Write the lesser number first.

(1 point)
Quadratic Equation Your Answer
1 (x+3)(x−1)=0
x =
and x =
2 (15x−5)(7+x)=0
x =
and x =
3 (13x−2)(x−34)=0
x =
and x =
4 30=x2+13x
x =
and x =

Answer 1

Let's solve each quadratic equation one by one:

1. **Equation:** \((x+3)(x−1)=0\)

To solve for \(x\):
- Set each factor to zero:
1. \(x + 3 = 0\) ⟹ \(x = -3\)
2. \(x - 1 = 0\) ⟹ \(x = 1\)

**Answer:** \(x = -3\) and \(x = 1\)

2. **Equation:** \((15x−5)(7+x)=0\)

To solve for \(x\):
- Set each factor to zero:
1. \(15x - 5 = 0\) ⟹ \(15x = 5\) ⟹ \(x = \frac{5}{15} = \frac{1}{3}\)
2. \(7 + x = 0\) ⟹ \(x = -7\)

**Answer:** \(x = -7\) and \(x = \frac{1}{3}\)

3. **Equation:** \((13x−2)(x−34)=0\)

To solve for \(x\):
- Set each factor to zero:
1. \(13x - 2 = 0\) ⟹ \(13x = 2\) ⟹ \(x = \frac{2}{13}\)
2. \(x - 34 = 0\) ⟹ \(x = 34\)

**Answer:** \(x = \frac{2}{13}\) and \(x = 34\)

4. **Equation:** \(30 = x^2 + 13x\)

Rearrange this to standard form:
\[ x^2 + 13x - 30 = 0 \]

Now, we can solve using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 1\), \(b = 13\), and \(c = -30\).

Calculate the discriminant:
\[
b^2 - 4ac = 13^2 - 4(1)(-30) = 169 + 120 = 289
\]

Now, plug this into the quadratic formula:
\[
x = \frac{-13 \pm \sqrt{289}}{2(1)} = \frac{-13 \pm 17}{2}
\]
Which gives:
1. \(x = \frac{4}{2} = 2\)
2. \(x = \frac{-30}{2} = -15\)

**Answer:** \(x = -15\) and \(x = 2\)

Here are the final answers summarized:

1. **Your Answer:** \(x = -3\) and \(x = 1\)
2. **Your Answer:** \(x = -7\) and \(x = \frac{1}{3}\)
3. **Your Answer:** \(x = \frac{2}{13}\) and \(x = 34\)
4. **Your Answer:** \(x = -15\) and \(x = 2\)