#1
For what positive numbers will the cube of a number exceed four times its square?
#2
For what positive numbers will the cube of a number be less than the number?
thanks
1. x^3 - 4x^2 > 0
x^2(x-4) > 0
x > 4
2. x^2 < x
x^2 - x < 0
x(x-2) < 0
0 < x < 2
#1 Well, let's have some cube fun! To find the positive numbers that make the cube of a number exceed four times its square, we can set up an inequality. Let's call the number "x". So, we have x^3 > 4x^2. To make things interesting, let's subtract 4x^2 from both sides to get x^3 - 4x^2 > 0. We can factor out an "x^2" from this inequality to get x^2(x - 4) > 0. Now, we have two factors - x^2 and (x - 4). To satisfy the inequality, either both factors must be greater than zero or both factors must be less than zero. Since we are looking for positive numbers, we can ignore the solution where both factors are less than zero. So, our condition is x > 0 and x - 4 > 0. Solving this, we get x > 4. So, any number greater than 4 will make the cube of that number exceed four times its square. Keep the clown-o-meter up! 🤡
#2 Now, let's dive into the world of shrinking cubes! To find the positive numbers that make the cube of a number less than the number itself, we can set up another inequality. Let's call the number "x" again. So, this time we have x^3 < x. Oh boy, things are getting interesting! To solve this, we need to consider two possibilities. First, x can be positive, which makes the inequality a bit tricky. Second, x can be negative, which surprisingly simplifies things. But since we are looking for positive numbers, let's focus on the first possibility. If x is positive, we can divide both sides of the inequality by x to get x^2 < 1. Since x is positive, its square will always be positive. Therefore, for the cube to be less than the number, we need x^2 to be strictly less than 1. Taking the square root of both sides, we get -1 < x < 1. So any number between -1 and 1 (excluding -1 and 1 itself) will satisfy the condition. Phew, that was a wild roller coaster ride! Keep the laughter rolling! 🎢🤡
To solve these problems, we can set up inequalities and solve them step by step.
#1:
Let's assume the number we are looking for is "x."
According to the problem, the cube of a number (x^3) should exceed four times its square (4x^2).
So we can write the inequality as: x^3 > 4x^2
To find the values of x that satisfy this inequality, we can rearrange it as follows: x^3 - 4x^2 > 0
Factor out an x^2: x^2(x - 4) > 0
We now have two conditions to consider:
1) x^2 > 0
2) (x - 4) > 0
For the first condition, x^2 > 0, any positive value of x will satisfy it.
For the second condition, (x - 4) > 0, we find that x > 4.
Combining these conditions, we can conclude that the values of x for which the cube of a number exceeds four times its square are x > 4.
#2:
Again, let's assume the number we are looking for is "x."
According to the problem, the cube of a number (x^3) should be less than the number (x).
So we can write the inequality as: x^3 < x
To find the values of x that satisfy this inequality, we can rearrange it as follows: x^3 - x < 0
To solve this inequality, we can factor out an x: x(x^2 - 1) < 0
Now we have two conditions to consider:
1) x < 0
2) (x^2 - 1) < 0
For the first condition, x < 0, any negative value of x will satisfy it.
For the second condition, (x^2 - 1) < 0, we find that -1 < x < 1.
Combining these conditions, we can conclude that the values of x for which the cube of a number is less than the number are: x < 0 and -1 < x < 1.
To solve these types of problems, we need to set up and solve an inequality.
#1: To find the positive numbers for which the cube of a number exceeds four times its square, we can express this inequality as:
x^3 > 4x^2
To solve this inequality, we can start by rearranging the terms:
x^3 - 4x^2 > 0
Next, we can factor out an 'x^2' term:
x^2(x - 4) > 0
Now, we have two factors to consider:
1. x^2 > 0: Since the square of any nonzero number is always positive, this factor is always satisfied.
2. x - 4 > 0: Solve this factor for x:
x > 4
Combining the results, we find that for any positive number greater than 4, the cube of the number will exceed four times its square.
Therefore, the solution to the first problem is x > 4.
#2: To find the positive numbers for which the cube of a number is less than the number, we can express this inequality as:
x^3 < x
To solve this inequality, we can rearrange the terms:
x^3 - x < 0
Again, we can factor out an 'x' term:
x(x^2 - 1) < 0
Now, we have two factors to consider:
1. x < 0: Since we are looking for positive numbers, this factor is not considered.
2. x^2 - 1 < 0: Solve this factor for x:
x^2 - 1 < 0
(x - 1)(x + 1) < 0
From this factorization, we can see that the inequality is satisfied when -1 < x < 1.
Therefore, the solution to the second problem is -1 < x < 1.