How do you solve #16x^2<15x+1#?

1 Answer
Dec 30, 2016

Answer:

The answer is #x in ] -1/16 , 1 [#

Explanation:

Let's rewrite the equation

#16x^2-15x-1<0#

Let #f(x)=16x^2-15x-1#

The domain of #f(x)# is #D_f(x)=RR#

We can factorise the RHS

#16x^2-15x-1=(16x+1)(x-1)#

Now we can establish the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-oo##color(white)(aaaaa)##-1/16##color(white)(aaaaa)##1##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##16x+1##color(white)(aaaaaaaa)##-##color(white)(aaaaaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-1##color(white)(aaaaaaaaaa)##-##color(white)(aaaaaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaaaa)##+##color(white)(aaaaaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)<0#, when #x in ] -1/16 , 1 [#