Question
Solve the inequality by factoring. 3x2−7x+2<0 Note: Enter your answer and SHOW ALL THE STEPS NECESSARY to solve this problem in the space provided. You will not receive credit without all work shown.
Answers
GPT-4o mini
To solve the inequality \(3x^2 - 7x + 2 < 0\) by factoring, follow these steps:
### Step 1: Factor the quadratic expression
We start with the expression \(3x^2 - 7x + 2\). We need to factor this quadratic.
We look for two numbers that multiply to \(3 \times 2 = 6\) (the product of the leading coefficient and the constant term) and add to \(-7\) (the coefficient of the linear term).
The two numbers that fulfill these conditions are \(-1\) and \(-6\) because:
\[
-1 \times -6 = 6 \quad \text{and} \quad -1 + (-6) = -7.
\]
### Step 2: Rewrite the middle term
We can rewrite \(-7x\) using \(-1x\) and \(-6x\):
\[
3x^2 - x - 6x + 2.
\]
### Step 3: Factor by grouping
Now we will group the terms:
\[
(3x^2 - x) + (-6x + 2).
\]
Next, factor out the common factors in each group:
\[
x(3x - 1) - 2(3x - 1).
\]
Now factor out the common binomial \((3x - 1)\):
\[
(3x - 1)(x - 2).
\]
### Step 4: Set up the inequality
Now we rewrite our inequality using the factored form:
\[
(3x - 1)(x - 2) < 0.
\]
### Step 5: Find the critical points
We set each factor to zero to find the critical points:
1. \(3x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{3}\),
2. \(x - 2 = 0 \quad \Rightarrow \quad x = 2\).
The critical points are \(x = \frac{1}{3}\) and \(x = 2\).
### Step 6: Test intervals
We will create intervals based on these critical points:
1. \( (-\infty, \frac{1}{3}) \)
2. \( \left(\frac{1}{3}, 2\right) \)
3. \( (2, \infty) \)
Now we will test each interval to see where the product \((3x - 1)(x - 2)\) is negative.
- **Interval 1: \( (-\infty, \frac{1}{3}) \)**
Choose \(x = 0\):
\[
(3(0) - 1)(0 - 2) = (-1)(-2) = 2 \quad \text{(positive)}.
\]
- **Interval 2: \( \left(\frac{1}{3}, 2\right) \)**
Choose \(x = 1\):
\[
(3(1) - 1)(1 - 2) = (3 - 1)(1 - 2) = (2)(-1) = -2 \quad \text{(negative)}.
\]
- **Interval 3: \( (2, \infty) \)**
Choose \(x = 3\):
\[
(3(3) - 1)(3 - 2) = (9 - 1)(3 - 2) = (8)(1) = 8 \quad \text{(positive)}.
\]
### Step 7: Collect results
The product \((3x - 1)(x - 2)\) is negative in the interval:
\[
\left(\frac{1}{3}, 2\right).
\]
### Step 8: Conclusion
Since the inequality is strict (<0), we do not include the endpoints.
Thus, the solution to the inequality \(3x^2 - 7x + 2 < 0\) is:
\[
\boxed{\left(\frac{1}{3}, 2\right)}.
\]
### Step 1: Factor the quadratic expression
We start with the expression \(3x^2 - 7x + 2\). We need to factor this quadratic.
We look for two numbers that multiply to \(3 \times 2 = 6\) (the product of the leading coefficient and the constant term) and add to \(-7\) (the coefficient of the linear term).
The two numbers that fulfill these conditions are \(-1\) and \(-6\) because:
\[
-1 \times -6 = 6 \quad \text{and} \quad -1 + (-6) = -7.
\]
### Step 2: Rewrite the middle term
We can rewrite \(-7x\) using \(-1x\) and \(-6x\):
\[
3x^2 - x - 6x + 2.
\]
### Step 3: Factor by grouping
Now we will group the terms:
\[
(3x^2 - x) + (-6x + 2).
\]
Next, factor out the common factors in each group:
\[
x(3x - 1) - 2(3x - 1).
\]
Now factor out the common binomial \((3x - 1)\):
\[
(3x - 1)(x - 2).
\]
### Step 4: Set up the inequality
Now we rewrite our inequality using the factored form:
\[
(3x - 1)(x - 2) < 0.
\]
### Step 5: Find the critical points
We set each factor to zero to find the critical points:
1. \(3x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{3}\),
2. \(x - 2 = 0 \quad \Rightarrow \quad x = 2\).
The critical points are \(x = \frac{1}{3}\) and \(x = 2\).
### Step 6: Test intervals
We will create intervals based on these critical points:
1. \( (-\infty, \frac{1}{3}) \)
2. \( \left(\frac{1}{3}, 2\right) \)
3. \( (2, \infty) \)
Now we will test each interval to see where the product \((3x - 1)(x - 2)\) is negative.
- **Interval 1: \( (-\infty, \frac{1}{3}) \)**
Choose \(x = 0\):
\[
(3(0) - 1)(0 - 2) = (-1)(-2) = 2 \quad \text{(positive)}.
\]
- **Interval 2: \( \left(\frac{1}{3}, 2\right) \)**
Choose \(x = 1\):
\[
(3(1) - 1)(1 - 2) = (3 - 1)(1 - 2) = (2)(-1) = -2 \quad \text{(negative)}.
\]
- **Interval 3: \( (2, \infty) \)**
Choose \(x = 3\):
\[
(3(3) - 1)(3 - 2) = (9 - 1)(3 - 2) = (8)(1) = 8 \quad \text{(positive)}.
\]
### Step 7: Collect results
The product \((3x - 1)(x - 2)\) is negative in the interval:
\[
\left(\frac{1}{3}, 2\right).
\]
### Step 8: Conclusion
Since the inequality is strict (<0), we do not include the endpoints.
Thus, the solution to the inequality \(3x^2 - 7x + 2 < 0\) is:
\[
\boxed{\left(\frac{1}{3}, 2\right)}.
\]