Question
Solve the quadratic inequality −2x2+8x−6≥0
by factoring.(1 point)
Given as an inequality, the solutions are
≤x≤
.
by factoring.(1 point)
Given as an inequality, the solutions are
≤x≤
.
Answers
Answered by
GPT-4o mini
To solve the quadratic inequality \(-2x^2 + 8x - 6 \geq 0\) by factoring, let's first rewrite it in a more manageable form:
1. Factor out \(-2\) from the left-hand side:
\[-2(x^2 - 4x + 3) \geq 0\]
2. Now, we can factor the quadratic expression \(x^2 - 4x + 3\):
\[x^2 - 4x + 3 = (x - 1)(x - 3)\]
3. Substitute this back into our inequality:
\[-2(x - 1)(x - 3) \geq 0\]
4. Next, we can divide both sides of the inequality by \(-2\). Remember that when dividing by a negative number, the inequality sign flips:
\[(x - 1)(x - 3) \leq 0\]
5. Now we need to find the critical points by setting the factors to zero:
\[x - 1 = 0 \Rightarrow x = 1\]
\[x - 3 = 0 \Rightarrow x = 3\]
6. These critical points divide the number line into intervals which we will test:
- \( (-\infty, 1) \)
- \( [1, 3] \)
- \( (3, \infty) \)
7. We test each interval:
- For \(x = 0\) in \((- \infty, 1)\):
\((0 - 1)(0 - 3) = (-1)(-3) = 3\) (positive, does not satisfy)
- For \(x = 2\) in \((1, 3)\):
\((2 - 1)(2 - 3) = (1)(-1) = -1\) (negative, satisfies)
- For \(x = 4\) in \((3, \infty)\):
\((4 - 1)(4 - 3) = (3)(1) = 3\) (positive, does not satisfy)
8. Thus, the solution to the inequality \((x - 1)(x - 3) \leq 0\) is within the interval \([1, 3]\).
Therefore, the final answer is:
\[1 \leq x \leq 3\]
This means the solutions are:
\[1 \leq x \leq 3\]
Or in inequality form:
\[\boxed{1 \leq x \leq 3}\]
1. Factor out \(-2\) from the left-hand side:
\[-2(x^2 - 4x + 3) \geq 0\]
2. Now, we can factor the quadratic expression \(x^2 - 4x + 3\):
\[x^2 - 4x + 3 = (x - 1)(x - 3)\]
3. Substitute this back into our inequality:
\[-2(x - 1)(x - 3) \geq 0\]
4. Next, we can divide both sides of the inequality by \(-2\). Remember that when dividing by a negative number, the inequality sign flips:
\[(x - 1)(x - 3) \leq 0\]
5. Now we need to find the critical points by setting the factors to zero:
\[x - 1 = 0 \Rightarrow x = 1\]
\[x - 3 = 0 \Rightarrow x = 3\]
6. These critical points divide the number line into intervals which we will test:
- \( (-\infty, 1) \)
- \( [1, 3] \)
- \( (3, \infty) \)
7. We test each interval:
- For \(x = 0\) in \((- \infty, 1)\):
\((0 - 1)(0 - 3) = (-1)(-3) = 3\) (positive, does not satisfy)
- For \(x = 2\) in \((1, 3)\):
\((2 - 1)(2 - 3) = (1)(-1) = -1\) (negative, satisfies)
- For \(x = 4\) in \((3, \infty)\):
\((4 - 1)(4 - 3) = (3)(1) = 3\) (positive, does not satisfy)
8. Thus, the solution to the inequality \((x - 1)(x - 3) \leq 0\) is within the interval \([1, 3]\).
Therefore, the final answer is:
\[1 \leq x \leq 3\]
This means the solutions are:
\[1 \leq x \leq 3\]
Or in inequality form:
\[\boxed{1 \leq x \leq 3}\]