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. only solutions numbers..,
Answers
GPT-4o mini
To solve the inequality \( 3x^2 - 7x + 2 < 0 \) by factoring, we follow these steps:
### Step 1: Factor the quadratic expression
We want to factor the quadratic \( 3x^2 - 7x + 2 \). We will look for two numbers that multiply to \( 3 \cdot 2 = 6 \) (the product of the coefficient of \( x^2 \) and the constant term) and add up to \( -7 \) (the coefficient of \( x \)).
The numbers that meet these criteria are \( -6 \) and \( -1 \) because:
- \( -6 \cdot -1 = 6 \)
- \( -6 + (-1) = -7 \)
Now we can rewrite the quadratic expression:
\[
3x^2 - 6x - 1x + 2
\]
Next, we group the terms:
\[
(3x^2 - 6x) + (-1x + 2)
\]
Factoring each group gives:
\[
3x(x - 2) - 1(x - 2)
\]
Now, we factor out \( (x - 2) \):
\[
(3x - 1)(x - 2)
\]
### Step 2: Set the factored expression to zero
To find the critical points, we set each factor equal to zero:
1. \( 3x - 1 = 0 \) gives \( x = \frac{1}{3} \)
2. \( x - 2 = 0 \) gives \( x = 2 \)
### Step 3: Determine the intervals for testing
The critical points divide the number line into three intervals:
1. \( (-\infty, \frac{1}{3}) \)
2. \( \left(\frac{1}{3}, 2\right) \)
3. \( (2, \infty) \)
### Step 4: Test points from each interval
Let's test a point from each interval to determine where the expression \( (3x - 1)(x - 2) < 0 \).
1. **For \( x = 0 \) (in \( (-\infty, \frac{1}{3}) \))**:
\[
(3(0) - 1)(0 - 2) = (-1)(-2) = 2 > 0
\]
The product is positive.
2. **For \( x = 1 \) (in \( \left(\frac{1}{3}, 2\right) \))**:
\[
(3(1) - 1)(1 - 2) = (3 - 1)(1 - 2) = (2)(-1) = -2 < 0
\]
The product is negative.
3. **For \( x = 3 \) (in \( (2, \infty) \))**:
\[
(3(3) - 1)(3 - 2) = (9 - 1)(3 - 2) = (8)(1) = 8 > 0
\]
The product is positive.
### Step 5: Combine the results
The product \( (3x - 1)(x - 2) < 0 \) is true in the interval \( \left(\frac{1}{3}, 2\right) \).
Therefore, 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 want to factor the quadratic \( 3x^2 - 7x + 2 \). We will look for two numbers that multiply to \( 3 \cdot 2 = 6 \) (the product of the coefficient of \( x^2 \) and the constant term) and add up to \( -7 \) (the coefficient of \( x \)).
The numbers that meet these criteria are \( -6 \) and \( -1 \) because:
- \( -6 \cdot -1 = 6 \)
- \( -6 + (-1) = -7 \)
Now we can rewrite the quadratic expression:
\[
3x^2 - 6x - 1x + 2
\]
Next, we group the terms:
\[
(3x^2 - 6x) + (-1x + 2)
\]
Factoring each group gives:
\[
3x(x - 2) - 1(x - 2)
\]
Now, we factor out \( (x - 2) \):
\[
(3x - 1)(x - 2)
\]
### Step 2: Set the factored expression to zero
To find the critical points, we set each factor equal to zero:
1. \( 3x - 1 = 0 \) gives \( x = \frac{1}{3} \)
2. \( x - 2 = 0 \) gives \( x = 2 \)
### Step 3: Determine the intervals for testing
The critical points divide the number line into three intervals:
1. \( (-\infty, \frac{1}{3}) \)
2. \( \left(\frac{1}{3}, 2\right) \)
3. \( (2, \infty) \)
### Step 4: Test points from each interval
Let's test a point from each interval to determine where the expression \( (3x - 1)(x - 2) < 0 \).
1. **For \( x = 0 \) (in \( (-\infty, \frac{1}{3}) \))**:
\[
(3(0) - 1)(0 - 2) = (-1)(-2) = 2 > 0
\]
The product is positive.
2. **For \( x = 1 \) (in \( \left(\frac{1}{3}, 2\right) \))**:
\[
(3(1) - 1)(1 - 2) = (3 - 1)(1 - 2) = (2)(-1) = -2 < 0
\]
The product is negative.
3. **For \( x = 3 \) (in \( (2, \infty) \))**:
\[
(3(3) - 1)(3 - 2) = (9 - 1)(3 - 2) = (8)(1) = 8 > 0
\]
The product is positive.
### Step 5: Combine the results
The product \( (3x - 1)(x - 2) < 0 \) is true in the interval \( \left(\frac{1}{3}, 2\right) \).
Therefore, the solution to the inequality \( 3x^2 - 7x + 2 < 0 \) is:
\[
\boxed{\left(\frac{1}{3}, 2\right)}
\]