x(6x+1)<15
can some one explain the problem and answer to me step by step please.. I missing the process
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
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
Sure! Let's break down the inequality step by step.
1. Start with the given inequality: x(6x+1)<15.
2. Simplify the expression inside the parentheses by distributing x: 6x^2 + x < 15.
3. Rearrange the inequality to have the expression on one side and zero on the other side: 6x^2 + x - 15 < 0.
4. To solve the inequality, we can find the critical points where the expression changes sign. To do this, set the quadratic expression equal to zero: 6x^2 + x - 15 = 0.
5. Factor the quadratic expression (if possible), or use the quadratic formula to solve for x. In this case, factoring is not straightforward, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a),
where a = 6, b = 1, and c = -15.
Plugging in the values, we get:
x = (-1 ± √(1^2 - 4*6*(-15))) / (2*6)
= (-1 ± √(1 + 360)) / 12
= (-1 ± √361) / 12
= (-1 ± 19) / 12
This gives us two possible solutions: x = (-1 + 19) / 12 and x = (-1 - 19) / 12.
6. Simplify the solutions:
x = 18 / 12
= 3/2,
or
x = -20 / 12
= -5/3.
7. Now, we need to plot these critical points (-5/3 and 3/2) on a number line.
8. Choose a test point close to each of the critical points and substitute it back into the inequality. Test for the sign of the expression. For example, take x = 0 (a value between -5/3 and 3/2) and substitute it into the inequality:
6x^2 + x - 15 < 0
6(0)^2 + 0 - 15 < 0
-15 < 0
Since -15 is indeed less than 0, the solution lies in the region between -5/3 and 3/2.
9. Finally, write the solution:
-5/3 < x < 3/2.
Therefore, the solution to the inequality x(6x+1) < 15 is -5/3 < x < 3/2.