Question $7$ of $30$ ( $1$ point) $\newline$ Express the solution in interval notation. $\newline$ $5 x^{2}+7 x-6>0$

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Multiply and divide rational expressions

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Q. Question

7

30

(

1

point)

\newline

Express the solution in interval notation.

\newline

5 x^{2}+7 x-6>0

Factor the quadratic equation: Question prompt: Express the solution in interval notation for the inequality $5x^{2}+7x-6 > 0$ .
Find zeros of the function: First, we need to factor the quadratic equation. So we're looking for two numbers that multiply to $-30$ $(5*-6)$ and add up to $7$ .
Test intervals around zeros: The numbers that work are $10$ and $-3$ because $10 \times -3 = -30$ and $10-3 = 7$ .
Check interval solutions: Now we can write the factored form of the quadratic: $(5x - 3)(x + 2) > 0$ .
Check interval solutions: Now we can write the factored form of the quadratic: 5x - 3)(x + 2) > 0\.Next, we find the zeros of the function by setting each factor equal to zero: \$5x - 3 = 0 and $x + 2 = 0$ .
Check interval solutions: Now we can write the factored form of the quadratic: 5x - 3)(x + 2) > 0\.Next, we find the zeros of the function by setting each factor equal to zero: \$5x - 3 = 0 and $x + 2 = 0$ .Solving these, we get $x = \frac{3}{5}$ and $x = -2$ .
Check interval solutions: Now we can write the factored form of the quadratic: 5x - 3)(x + 2) > 0\.Next, we find the zeros of the function by setting each factor equal to zero: \$5x - 3 = 0 and $x + 2 = 0$ .Solving these, we get $x = \frac{3}{5}$ and $x = -2$ .Now we'll test intervals around the zeros to see where the inequality is greater than zero. The intervals are $\left(-\infty, -2\right)$ , $\left(-2, \frac{3}{5}\right)$ , and $\left(\frac{3}{5}, \infty\right)$ .
Check interval solutions: Now we can write the factored form of the quadratic: $(5x - 3)(x + 2) > 0$ .Next, we find the zeros of the function by setting each factor equal to zero: $5x - 3 = 0$ and $x + 2 = 0$ .Solving these, we get $x = \frac{3}{5}$ and $x = -2$ .Now we'll test intervals around the zeros to see where the inequality is greater than zero. The intervals are $(-\infty, -2)$ , $(-2, \frac{3}{5})$ , and $(\frac{3}{5}, \infty)$ .Pick test points from each interval, like $x = -3$ , $x = 0$ , and $5x - 3 = 0$ $0$ , and plug them into the factored inequality to see if it's true.
Check interval solutions: Now we can write the factored form of the quadratic: 5x - 3)(x + 2) > 0\.Next, we find the zeros of the function by setting each factor equal to zero: \$5x - 3 = 0 and $x + 2 = 0$ .Solving these, we get $x = \frac{3}{5}$ and $x = -2$ .Now we'll test intervals around the zeros to see where the inequality is greater than zero. The intervals are $\left(-\infty, -2\right)$ , $\left(-2, \frac{3}{5}\right)$ , and $\left(\frac{3}{5}, \infty\right)$ .Pick test points from each interval, like $x = -3$ , $x = 0$ , and $x = 1$ , and plug them into the factored inequality to see if it's true.For $x = -3$ , we get $x + 2 = 0$ $1$ , which simplifies to $x + 2 = 0$ $2$ . This is true, so $\left(-\infty, -2\right)$ is part of the solution.
Check interval solutions: Now we can write the factored form of the quadratic: 5x - 3)(x + 2) > 0\.Next, we find the zeros of the function by setting each factor equal to zero: \$5x - 3 = 0 and $x + 2 = 0$ .Solving these, we get $x = \frac{3}{5}$ and $x = -2$ .Now we'll test intervals around the zeros to see where the inequality is greater than zero. The intervals are $\left(-\infty, -2\right)$ , $\left(-2, \frac{3}{5}\right)$ , and $\left(\frac{3}{5}, \infty\right)$ .Pick test points from each interval, like $x = -3$ , $x = 0$ , and $x = 1$ , and plug them into the factored inequality to see if it's true.For $x = -3$ , we get $x + 2 = 0$ $1$ , which simplifies to $x + 2 = 0$ $2$ . This is true, so $\left(-\infty, -2\right)$ is part of the solution.For $x = 0$ , we get $x + 2 = 0$ $5$ , which simplifies to $x + 2 = 0$ $6$ . This is false, so $\left(-2, \frac{3}{5}\right)$ is not part of the solution.
Check interval solutions: Now we can write the factored form of the quadratic: 5x - 3)(x + 2) > 0\.Next, we find the zeros of the function by setting each factor equal to zero: \$5x - 3 = 0 and $x + 2 = 0$ .Solving these, we get $x = \frac{3}{5}$ and $x = -2$ .Now we'll test intervals around the zeros to see where the inequality is greater than zero. The intervals are $\left(-\infty, -2\right)$ , $\left(-2, \frac{3}{5}\right)$ , and $\left(\frac{3}{5}, \infty\right)$ .Pick test points from each interval, like $x = -3$ , $x = 0$ , and $x = 1$ , and plug them into the factored inequality to see if it's true.For $x = -3$ , we get $x + 2 = 0$ $1$ , which simplifies to $x + 2 = 0$ $2$ . This is true, so $\left(-\infty, -2\right)$ is part of the solution.For $x = 0$ , we get $x + 2 = 0$ $5$ , which simplifies to $x + 2 = 0$ $6$ . This is false, so $\left(-2, \frac{3}{5}\right)$ is not part of the solution.For $x = 1$ , we get $x + 2 = 0$ $9$ , which simplifies to $x = \frac{3}{5}$ $0$ . This is true, so $\left(\frac{3}{5}, \infty\right)$ is part of the solution.
Check interval solutions: Now we can write the factored form of the quadratic: $(5x - 3)(x + 2) > 0$ .Next, we find the zeros of the function by setting each factor equal to zero: $5x - 3 = 0$ and $x + 2 = 0$ .Solving these, we get $x = \frac{3}{5}$ and $x = -2$ .Now we'll test intervals around the zeros to see where the inequality is greater than zero. The intervals are $(-\infty, -2)$ , $(-2, \frac{3}{5})$ , and $(\frac{3}{5}, \infty)$ .Pick test points from each interval, like $x = -3$ , $x = 0$ , and $5x - 3 = 0$ $0$ , and plug them into the factored inequality to see if it's true.For $x = -3$ , we get $5x - 3 = 0$ $2$ , which simplifies to $5x - 3 = 0$ $3$ . This is true, so $(-\infty, -2)$ is part of the solution.For $x = 0$ , we get $5x - 3 = 0$ $6$ , which simplifies to $5x - 3 = 0$ $7$ . This is false, so $(-2, \frac{3}{5})$ is not part of the solution.For $5x - 3 = 0$ $0$ , we get $x + 2 = 0$ $0$ , which simplifies to $x + 2 = 0$ $1$ . This is true, so $(\frac{3}{5}, \infty)$ is part of the solution.The solution in interval notation is $x + 2 = 0$ $3$ .

Question 777 of 303030 (111 point)\newlineExpress the solution in interval notation.\newline5x2+7x−6>0 5 x^{2}+7 x-6>0 5x2+7x−6>0

Question $7$ of $30$ ( $1$ point) $\newline$ Express the solution in interval notation. $\newline$ $5 x^{2}+7 x-6>0$