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.
Find the number of integers that satisfy $x(3x-4) \le \frac{6x^2 - 3x + 5}{5}$.
5 answers
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
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 $\boxed{\left[\frac{17-\sqrt{469}}{18}, \frac{17+\sqrt{469}}{18}\right]}$.
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.
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 $\boxed{3}$.
1 answer