Friday

January 30, 2015

January 30, 2015

Posted by **ladybug** on Saturday, September 22, 2012 at 4:24pm.

16(x^2-1) >16x

the solution set is? or there is no solution.

- pre-calculus -
**Anonymous**, Saturday, September 22, 2012 at 4:40pmThis 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

- pre-calculus -
**Steve**, Saturday, September 22, 2012 at 4:54pmthe 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.

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