solve the following inequality
16(x^2-1) >16x
the solution set is? or there is no solution.
This problem can be solved exactly like a quadratic equation would be:
16(x^2-1) > 16x
16x^2 - 16 > 16x
16x^2 - 16x - 16 > 0
a b c
You use the quadratic formula:
x = (-b +/- sqrt(b^2 - 4ac))/2a
the problem here is that after having solved the equation, you now know where the graph crosses the x-axis. You want to know where
16(x^2-1) > 16x
x^2 - 1 > x
x^2 - x - 1 > 0
Now from what you know about parabolas, it should clear that the graph is above the x-axis everywhere except between the roots, which you found using the method described above.
To solve the inequality 16(x^2-1) > 16x, we can start by simplifying the expression on the left side:
16(x^2-1) > 16x
16x^2 - 16 > 16x
Next, we can move all the terms to one side to get a quadratic equation:
16x^2 - 16 - 16x > 0
16x^2 - 16x - 16 > 0
Now, we need to find the critical points, where the inequality changes. We can do this by setting the quadratic equation equal to zero and solving for x:
16x^2 - 16x - 16 = 0
To simplify the equation, let's divide all terms by 16:
x^2 - x - 1 = 0
Now, we can solve this quadratic equation by factoring, using the quadratic formula, or completing the square. In this case, the quadratic equation does not factor easily, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = -1, and c = -1. Plugging these values into the quadratic formula, we get:
x = (-(-1) ± √((-1)^2 - 4(1)(-1))) / (2(1))
x = (1 ± √(1 + 4)) / 2
x = (1 ± √5) / 2
Now we have two critical points: (1 + √5) / 2 and (1 - √5) / 2. We can use these points to divide the real number line into three intervals.
Interval 1: x < (1 - √5) / 2
Interval 2: (1 - √5) / 2 < x < (1 + √5) / 2
Interval 3: x > (1 + √5) / 2
Now, we can choose a test point in each interval and substitute it into the original inequality to determine if it is true or false.
For Interval 1, let's choose x = 0:
16(0^2 - 1) > 16(0)
16(-1) > 0
-16 > 0
Since -16 is not greater than 0, the inequality is false in Interval 1.
For Interval 2, let's choose x = 1:
16(1^2 - 1) > 16(1)
16(0) > 16
0 > 16
Since 0 is not greater than 16, the inequality is false in Interval 2.
For Interval 3, let's choose x = 2:
16(2^2 - 1) > 16(2)
16(3) > 32
48 > 32
Since 48 is greater than 32, the inequality is true in Interval 3.
Therefore, the solution to the inequality 16(x^2-1) > 16x is x > (1 + √5) / 2.