#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 To find the positive numbers for which the cube of a number exceeds four times its square, we can set up an equation.
Let's assume the number is represented by "x".
The cube of the number is given by x^3, and four times its square is 4x^2.
So, we need to solve the inequality x^3 > 4x^2.
Simplifying the inequality, we get x^3 - 4x^2 > 0.
To solve this inequality, we can factor out an "x^2" term: x^2(x - 4) > 0.
The inequality is satisfied when either the factors x^2 and (x - 4) are both positive or both negative.
If x^2 > 0 (x is positive), then (x - 4) should also be > 0.
So, x > 0 and x - 4 > 0. Solving these inequalities, we get x > 4.
Therefore, the positive values for which the cube of a number exceeds four times its square are x > 4.
#2 To find the positive numbers for which the cube of a number is less than the number, we can set up an equation.
Let's assume the number is represented by "x".
The cube of the number is given by x^3, and the number itself is denoted by x.
So, we need to solve the inequality x^3 < x.
To solve this inequality, we can factor out an "x" term: x(x^2 - 1) < 0.
The inequality is satisfied when either the factors x and (x^2 - 1) are both positive or both negative.
If x > 0, then (x^2 - 1) should also be > 0.
So, x > 0 and x^2 - 1 > 0. Simplifying the second inequality, we get x > 1 or x < -1.
Therefore, the positive values for which the cube of a number is less than the number are 0 < x < 1.
To find the answers to these questions, we can solve the given equations algebraically.
#1: To determine when the cube of a number exceeds four times its square, we can set up the equation:
x^3 > 4x^2
Here, x represents the number we are trying to find. To solve this inequality, we can simplify it by dividing both sides by x^2 (since x^2 is always positive, we don't need to worry about reversing the inequality):
x > 4
So, any positive number greater than 4 will satisfy the condition. Therefore, the answer to the first question is: for any positive number greater than 4.
#2: To determine when the cube of a number is less than the number, we can set up the equation:
x^3 < x
Again, x represents the number we are trying to find. To solve this inequality, we can rearrange it by subtracting x from both sides:
x^3 - x < 0
Now, we can factor out x:
x(x^2 - 1) < 0
The expression x^2 - 1 can be factored as (x - 1)(x + 1). Now, we have two factors of the inequality:
x(x - 1)(x + 1) < 0
To solve this, we can use a sign chart, where we test the sign of the expression for different intervals of x. We find that the inequality is satisfied when x is between -1 and 0, and also when x is greater than 1. Therefore, the answer to the second question is: for any positive number greater than 1, but less than -1.
It is important to note that solving algebraic inequalities requires a good understanding of algebraic concepts such as factoring and solving quadratic equations.