# math

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x(6x+1)<15
can some one explain the problem and answer to me step by step please.. I missing the process

• math -

x(6x+1) < 15
6x^2 + x - 15 < 0
now you have a parabola. You need to find where it crosses the x-axis, and between those two points, it will lie below the x-axis.

(2x-3)(3x+5) < 0

so, the roots are at x = -5/3 and 3/2

Between those two values, the parabola lies below the x-axis.

As a check, let's try the original conditions.

x(6x+1) < 15
at x = -5/3, x(6x+1) = 15
at x = 3/2, x(6x+1) = 15
Now check something in between, say x = -1,0,1
f(x) = 5,0,7 all < 15.

How about a few outside -5/3 < x < 3/2 ?
f(-2) = 22 > 15
f(2) = 26 > 15

Take a look here:

http://www.wolframalpha.com/input/?i=x%286x%2B1%29+%3C+15

• math -

Initial Problem
x(6x + 1) < 15

Simplify left side.
6x^2 + x < 15

Move 15 to left side.
6x^2 + x - 15 < 0

Factor.
(2x - 3)(3x + 5) < 0

Now, x MUST be BETWEEN the LESSER value and the GREATER value. (it is between because this is a "LESS THAN" inequality)
-5/3 < x < 3/2

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