Please please help!!!!!
Solve the inequality and write the solution set in interval notation
4x^3 -6x^2<0
2x^2 (2x-3)<0
2x-3<0
x= 3/2
Now I'm not sure about this
2x^2<0
x^2<1/2
sq root of x^2<sq root of 1/2
x=.7071
Right?
When I plug it in the calculator it says
X=0
How?
4x^3 - 6x^2 < 0
2x^2 (2x-3) < 0
You should know the general shape of this graph. It has a double root at x=0 and it crosses the x-axis at x=3/2.
So, the graph is below the x-axis for all values of x less than 3/2, except x=0.
2x-3 < 0 is an inequality.
You cannot change it to an equation: x = 3/2
2x-3 < 0
2x < 3
x < 3/2
--------------------------------------
If 2x^2 < 0
you cannot say x^2 < 1/2
0/2 = 0
HINT: x^2 is always >=0, so 2x^2 is always >= 0.
Always
(a) check your answer in the original expression
(b) THINK about your answer in the context of the problem!
x≤−1 or x≥7
To solve the inequality 4x^3 - 6x^2 < 0, you can follow these steps:
Step 1: Factor out the common factor, which is 2x^2:
2x^2(2x - 3) < 0
Step 2: Set each factor equal to zero and solve for x separately:
2x^2 = 0 --> x = 0
2x - 3 = 0 --> x = 3/2
Step 3: Create a number line and plot the critical points x = 0 and x = 3/2:
-----0----3/2-----
Step 4: Test intervals between and outside the critical points. Choose any value from each interval and substitute it into the original inequality to determine if it satisfies the inequality or not.
Interval 1: (-∞, 0)
Choose x = -1:
4(-1)^3 - 6(-1)^2 ---> -4 - 6 < 0, which is true.
So, this interval satisfies the inequality.
Interval 2: (0, 3/2)
Choose x = 1:
4(1)^3 - 6(1)^2 ---> 4 - 6 < 0, which is false.
So, this interval does not satisfy the inequality.
Interval 3: (3/2, +∞)
Choose x = 2:
4(2)^3 - 6(2)^2 ---> 32 - 24 < 0, which is false.
So, this interval does not satisfy the inequality.
Step 5: Write the solution in interval notation:
The solution set for 4x^3 - 6x^2 < 0 is (-∞, 0).
For the second inequality, 2x^2(2x-3) < 0, you can follow the same steps as above:
Step 1: Factor out the common factor, which is 2x^2:
2x^2(2x - 3) < 0
Step 2: Set each factor separately to zero and solve for x:
2x^2 = 0 --> x = 0
2x - 3 = 0 --> x = 3/2
Step 3: Create a number line and plot the critical points x = 0 and x = 3/2:
-----0----3/2-----
Step 4: Test intervals between and outside the critical points. Choose any value from each interval and substitute it into the original inequality to determine if it satisfies the inequality or not.
Interval 1: (-∞, 0)
Choose x = -1:
2(-1)^2(2(-1) - 3) < 0, which is false.
So, this interval does not satisfy the inequality.
Interval 2: (0, 3/2)
Choose x = 1:
2(1)^2(2(1) - 3) < 0, which is true.
So, this interval satisfies the inequality.
Interval 3: (3/2, +∞)
Choose x = 2:
2(2)^2(2(2) - 3) < 0, which is false.
So, this interval does not satisfy the inequality.
Step 5: Write the solution in interval notation:
The solution set for 2x^2(2x-3) < 0 is (0, 3/2).
Now, regarding your confusion about the value of x when you plug it into the calculator, it seems like there might be an error.
For the equation x^2 < 1/2, when you take the square root, you should consider both positive and negative square roots. So:
sqrt(x^2) < sqrt(1/2)
|x| < sqrt(1/2)
Taking the square root of 1/2 gives you √(1/2) ≈ 0.7071. However, when you take the absolute value of x, you should consider both positive and negative roots. So the solution set would be x < 0.7071 or x > -0.7071.
So, if you plug in x = 0.7071 or x = -0.7071 into the original equation x^2 < 1/2, it should satisfy the inequality. However, when you plug these values into the calculator, you might encounter rounding errors, resulting in x = 0.
In conclusion, the solution to x^2 < 1/2 is x < 0.7071 or x > -0.7071.