Solve 3𝑥^2 − 8𝑥 + 4 ≤ 0 using the number-line method.
To solve 3𝑥^2 − 8𝑥 + 4 ≤ 0 using the number-line method, we need to find the values of 𝑥 that make the inequality true.
First, let's find the values of 𝑥 that make the inequality equal to zero:
3𝑥^2 − 8𝑥 + 4 = 0
To factor the quadratic equation, we can use the quadratic formula: 𝑥 = (-𝑏 ± √(𝑏^2 - 4𝑎𝑐))/(2𝑎)
Using this formula, we have:
𝑥 = (-(-8) ± √((-8)^2 - 4(3)(4)))/(2(3))
𝑥 = (8 ± √(64 - 48))/(6)
𝑥 = (8 ± √(16))/(6)
𝑥 = (8 ± 4)/(6)
This gives us two values for 𝑥: 𝑥 = 2 and 𝑥 = 1/3.
Now, let's plot these values on a number line:
----(1/3)---(2)----
Next, we need to determine the sign of the inequality in each interval:
When 𝑥 < 1/3:
Plug in a test value less than 1/3 into the inequality, such as 𝑥 = 0:
3(0)^2 - 8(0) + 4 ≤ 0
0 - 0 + 4 ≤ 0
4 ≤ 0
This is false, so the inequality is not true for 𝑥 < 1/3.
When 1/3 < 𝑥 < 2:
Plug in a test value between 1/3 and 2 into the inequality, such as 𝑥 = 1:
3(1)^2 - 8(1) + 4 ≤ 0
3 - 8 + 4 ≤ 0
-1 ≤ 0
This is true, so the inequality is true for 1/3 < 𝑥 < 2.
When 𝑥 > 2:
Plug in a test value greater than 2 into the inequality, such as 𝑥 = 3:
3(3)^2 - 8(3) + 4 ≤ 0
27 - 24 + 4 ≤ 0
7 ≤ 0
This is false, so the inequality is not true for 𝑥 > 2.
Now we can write the solution using the number-line method:
1/3 < 𝑥 < 2