Find the number of integers that satisfy $x(3x-4) \le \frac{6x^2 - 3x + 5}{5}$.
AAAaannndd the bot gets it wrong yet again!
9x^2 - 17x - 5 does not factor as (3x-1)(3x-5)
9x^2 - 17x - 5 <= 0
The roots of 9x^2 - 17x - 5 = 0 are
(17±√469)/18
Since the parabola opens upward, we want the interval
(17-√469)/18 <= x <= (17+√469)/18
You are correct, and I apologize for the error. To solve the inequality $9x^2 - 17x - 5 \leq 0$, we need to find the interval over which the quadratic is less than or equal to zero.
The roots of the quadratic are $\frac{17 \pm \sqrt{469}}{18}$. Since the coefficient of $x^2$ is positive, the graph of the quadratic opens upwards.
To determine the sign of $9x^2 - 17x - 5$, we can test a value in each of the intervals defined by the roots. We can choose the values $x = 0$ and $x = 1$.
For $x = 0$, we have $9(0)^2 - 17(0) - 5 < 0$, so the quadratic is negative in the interval $(17-\sqrt{469})/18 < x < (17+\sqrt{469})/18$.
For $x = 1$, we have $9(1)^2 - 17(1) - 5 > 0$, so the quadratic is positive outside the interval $(17-\sqrt{469})/18 < x < (17+\sqrt{469})/18$.
Therefore, the inequality is satisfied in the interval $\boxed{\left[\frac{17-\sqrt{469}}{18}, \frac{17+\sqrt{469}}{18}\right]}$.
my bad - I did not notice we are looking for integer solutions.
Since those are 0,1,2 there are three integer solutions.
Yes, you are correct. We are looking for integer solutions, and the only integers that satisfy the inequality are $0, 1,$ and $2$. Thus, the number of integers that satisfy the inequality is $\boxed{3}$.
To solve this inequality, we can start by simplifying both sides of the equation:
$x(3x-4) \le \frac{6x^2 - 3x + 5}{5}$
After simplifying, let's bring everything to one side of the inequality:
$5x(3x-4) - (6x^2 - 3x + 5) \le 0$
Next, let's expand and combine like terms:
$15x^2 - 20x - 6x^2 + 3x - 5 \le 0$
Combine the $x^2$ terms and the x terms:
$9x^2 - 17x - 5 \le 0$
To continue solving this quadratic inequality, we need to find the values of x that satisfy $9x^2 - 17x - 5 \le 0$. To do that, we can find the zeros by factoring or using the quadratic formula.
Factoring $9x^2 - 17x - 5$ yields:
$(3x+1)(3x-5) \le 0$
Now we have the equation in factored form. To find the values of x that satisfy the inequality, we can use a sign chart.
| Interval | $(3x+1)(3x-5)$<0 |
|-------------|----------------------|
| $x < -\frac{1}{3}$ | - |
| $-1<x<\frac{5}{3}$ | + |
| $x > \frac{5}{3}$ | - |
From the sign chart, we see that the inequality is satisfied when $-1 < x < \frac{5}{3}$.
So, the number of integers that satisfy the inequality is the number of integers within the interval $(-1, \frac{5}{3})$.
To find the number of integers that satisfy the given inequality, we first need to simplify it.
Let's start by multiplying both sides of the inequality by 5 to clear the denominator:
$5x(3x-4) \le 6x^2 - 3x + 5$
Simplifying the left side:
$15x^2 - 20x \le 6x^2 - 3x + 5$
Combining like terms on the right side:
$15x^2 - 20x \le 6x^2 - 3x + 5$
Now, let's rearrange the terms to bring all terms to one side of the inequality:
$15x^2 - 6x^2 - 20x + 3x \le 5$
Simplifying further:
$9x^2 - 17x \le 5$
To solve this quadratic inequality, we set it equal to zero:
$9x^2 - 17x - 5 \le 0$
Next, we can find the zeros by factoring the quadratic equation:
$(3x +1)(3x - 5) \le 0$
The zeros are $x = -\frac{1}{3}$ and $x = \frac{5}{3}$.
Now, we can create a sign chart to determine the sign of the expression in each interval:
Interval 1: $x < -\frac{1}{3}$
Interval 2: $-\frac{1}{3} < x < \frac{5}{3}$
Interval 3: $x > \frac{5}{3}$
Now, we choose a test point from each interval and evaluate the inequality.
For Interval 1, let's choose $x = -1$:
$(3(-1) + 1)(3(-1) - 5) \le 0$
$(-2)(-8) \le 0$
$16 \le 0$ is false
For Interval 2, let's choose $x = 0$:
$(3(0) + 1)(3(0) - 5) \le 0$
$(1)(-5) \le 0$
$-5 \le 0$ is true
For Interval 3, let's choose $x = 2$:
$(3(2) + 1)(3(2) - 5) \le 0$
$(7)(1) \le 0$
$7 \le 0$ is false
From the sign chart and evaluating the inequality at the test points, we find that the solution set for the inequality is $-\frac{1}{3} \le x < \frac{5}{3}$.
Since the question asks for the number of integers that satisfy the inequality, we need to count the number of integers in this solution set.
To do this, we observe that the integer solutions are $x = -1, 0, 1$. Thus, there are 3 integers that satisfy the given inequality.
Multiplying through by 5, we have \[5x(3x - 4) \le 6x^2 - 3x + 5.\]Expanding, we get \[15x^2 - 20x \le 6x^2 - 3x + 5,\]or \[9x^2 - 17x + 5 \le 0.\]This quadratic factors as \[(3x - 1)(3x - 5) \le 0,\]so we have either $3x - 1 \le 0$ and $3x - 5 \ge 0,$ or $3x - 1 \ge 0$ and $3x - 5 \le 0.$
In the first case, $x \le \frac{1}{3}$ and $x \ge \frac{5}{3},$ which is contradictions, so there are no solutions in this case.
In the second case, $x \ge \frac{1}{3}$ and $x \le \frac{5}{3},$ which leads to $\frac{1}{3} \le x \le \frac{5}{3}.$ The integers in this interval are 1 and 2, so the number of solutions is $\boxed{2}.$
Note: To finish the problem, we must check that the quadratic actually satisfies case 2. But we see that $3 \cdot \frac{1}{3} - 1 = 0$ and $3 \cdot \frac{5}{3} - 5 = 0,$ so case 2 is valid.