For what positive numbers will the square of a number exceed twice the number?
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To determine the positive numbers for which the square of a number exceeds twice the number, we can set up an inequality and solve it step-by-step.
Let's assume the number is "x."
The square of the number can be represented as x^2, and twice the number can be represented as 2x.
We can write the inequality as:
x^2 > 2x
To solve this inequality, we need to bring all terms to one side of the inequality sign. Therefore, we subtract 2x from both sides:
x^2 - 2x > 0
Now, we have a quadratic inequality. To solve it, we find the x-intercepts of the corresponding quadratic equation:
x(x - 2) > 0
To find the values of x that satisfy this inequality, we need to consider the sign of each term and how they affect the inequality.
1. If the sign of the product of x and x - 2 is positive (+), it means that both terms have the same sign (either both positive or both negative).
2. If the sign of the product is negative (-), it means that one term is positive while the other is negative.
Now we can analyze each case:
Case 1: Both terms are positive (+):
To satisfy the inequality x(x - 2) > 0, we have the following possibilities:
a) x > 0 and x - 2 > 0
This implies x > 2, which means x must be greater than 2.
b) x < 0 and x - 2 < 0
This implies x < 0, which means x must be less than 0.
However, in our case, we are only considering positive numbers. Therefore, case 1 doesn't yield any relevant values.
Case 2: One term is positive and the other is negative:
To satisfy the inequality x(x - 2) > 0, we have the following possibility:
a) x > 0 and x - 2 < 0
In this case, x > 2, which means x must be greater than 2.
Thus, the positive numbers for which the square of a number exceeds twice the number are x > 2.
To summarize:
The positive numbers for which the square of a number exceeds twice the number are x > 2.
To find the positive numbers for which the square of a number exceeds twice the number, we can translate the problem into an equation and then solve it algebraically.
Let's assume the number we are looking for is 'x'. According to the problem, the square of 'x' should exceed twice 'x'. We can write this inequality as:
x^2 > 2x
To solve this equation, we need to rearrange it to one side and find the values of 'x' that satisfy it.
x^2 - 2x > 0
Now, we have a quadratic inequality. To solve it, we factorize the equation:
x(x - 2) > 0
Now we have two key points to consider: x = 0 and x - 2 = 0, which results in x = 2.
We can construct a sign chart to determine the intervals of 'x' that satisfy the inequality:
| - | 0 | 2 | + |
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| (-) | (0) | (+) | (+) |
From the sign chart, it is evident that the inequality is satisfied for x < 0 and x > 2. This means that the positive numbers for which the square of a number exceeds twice the number are x > 2.
Therefore, any positive number greater than 2 will satisfy the given condition.
n^2>2n
n^2-2n>0
n(n-2)>0
So, n must be 3 or greater for the second term to be positive. If it is not positive, the inequality is not true.