Solve the polynomial equation and graphy it on a number line. Express the solution set and interval notation. and which brackets to use either () or []
15X^2 < 14x + 1
This is an inequation, not an equation.
take everything to one side
15x^2 - 14x - 1 < 0, it factors
(15x+1)(x-1) < 0
If this had been the equation
y = (15x+1)(x-1) it would have been a parabola opening upwards with x-intercepts of -1/15 and 1
Between those two values the graph would have been below the x-axis
so the solution set is -1/5 < x < 1
so the solution set is -1/15 < x < 1
To solve the polynomial equation and graph it on a number line, let's start by rearranging the equation to bring all terms to one side:
15x^2 - 14x - 1 < 0
Now, we can find the solutions by factoring, completing the square, or using the quadratic formula. In this case, factoring would be difficult, completing the square might be too lengthy, so let's use the quadratic formula to find the solutions:
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions can be found using:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 15, b = -14, and c = -1. Plugging these values into the quadratic formula:
x = (-(-14) ± √((-14)^2 - 4(15)(-1))) / (2(15))
Simplifying:
x = (14 ± √(196 + 60)) / 30
x = (14 ± √256) / 30
x = (14 ± 16) / 30
This gives us two possible solutions:
x1 = (14 + 16) / 30 = 30 / 30 = 1
x2 = (14 - 16) / 30 = -2 / 30 = -1/15
Now, let's graph the equation on a number line. Here's how to do it:
1. Draw a horizontal line and label it with a number line scale.
2. Mark the critical points on the number line. In this case, we have x = -1/15 and x = 1.
3. Choose a test value within each of the intervals created.
4. Substitute the test values into the original inequality and determine whether the inequality is true or false.
5. Mark the intervals where the inequality is true on the number line.
Based on the solutions we found, the number line can be divided into three intervals: (-∞, -1/15), (-1/15, 1), and (1, +∞). Now, we need to test a value from each interval.
For the interval (-∞, -1/15), we can choose a value such as x = -1.
Substituting x = -1 into the inequality:
15(-1)^2 - 14(-1) - 1 < 0
15 - (-14) - 1 < 0
15 + 14 - 1 < 0
28 < 0
Since 28 is not less than 0, this interval does not satisfy the inequality.
For the interval (-1/15, 1), we can choose a value such as x = 0.
Substituting x = 0 into the inequality:
15(0)^2 -14(0) - 1 < 0
0 - 0 - 1 < 0
-1 < 0
Since -1 is indeed less than 0, this interval satisfies the inequality.
Finally, for the interval (1, +∞), we can choose a value such as x = 2.
Substituting x = 2 into the inequality:
15(2)^2 - 14(2) - 1 < 0
60 - 28 - 1 < 0
31 < 0
Again, 31 is not less than 0, so this interval does not satisfy the inequality.
Now, we can express the solution set using interval notation and the appropriate brackets:
(-1/15, 1]