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
solve the following inequality
16(x^2-1) >16x
the solution set is? or there is no solution.
2 answers
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.
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.