Question

question 3 (5 points)
what is the solution to the inequality \\(x^2 + 2x \le 24\\)?
a) \\(-6 < x < 4\\)
b) \\(x < 4\\) or \\(x > -6\\)
c) \\(-6 \le x \le 4\\)
d) \\(x \le -6\\) or \\(x \ge 4\\)

question 4 (5 points)
solve the system of equations using substitution or graphing.
\\(y = -x^2 + 4x + 5\\)
\\(y = x + 1\\)
a) \\((-1, 0)\\) and \\((6, 7)\\)
b) \\((1, 2)\\) and \\((6, 7)\\)
c) \\((1, 2)\\) and \\((4, 5)\\)
d) \\((-1, 0)\\) and \\((4, 5)\\)

Explanation:

Solve the quadratic inequality

We want to solve the inequality:

$$x^2 + 2x \le 24$$

First, we rearrange it into standard form by subtracting \(24\) from both sides:

$$x^2 + 2x - 24 \le 0$$

Next, we find the boundary points by factoring the quadratic expression:

$$(x + 6)(x - 4) \le 0$$

The roots of the equation are \(x = -6\) and \(x = 4\). Since the inequality is \(\le 0\), the solution is the interval between and including these boundary points:

$$-6 \le x \le 4$$

This corresponds to option C.

Solve the system of equations

We are given the system:

$$
LATEXBLOCK0
$$

We substitute the second equation into the first:

$$x + 1 = -x^2 + 4x + 5$$

Rearranging all terms to one side to form a quadratic equation:

$$x^2 - 3x - 4 = 0$$

Factoring the quadratic equation:

$$(x - 4)(x + 1) = 0$$

This gives the \(x\)-values:

$$x = 4 \quad \text{or} \quad x = -1$$

Now, we find the corresponding \(y\)-values using \(y = x + 1\):

• For \(x = 4\): \(y = 4 + 1 = 5\), giving the point \((4, 5)\).
• For \(x = -1\): \(y = -1 + 1 = 0\), giving the point \((-1, 0)\).

The solutions are \((-1, 0)\) and \((4, 5)\), which corresponds to option D.

Answer:

Question 3

• A) \(-6 < x < 4\)
• B) \(x < 4 \text{ or } x > -6\)
• C) \(-6 \le x \le 4\) (Correct answer)
• D) \(x \le -6 \text{ or } x \ge 4\)

Question 4

• A) \((-1, 0)\) and \((6, 7)\)
• B) \((1, 2)\) and \((6, 7)\)
• C) \((1, 2)\) and \((4, 5)\)
• D) \((-1, 0)\) and \((4, 5)\) (Correct answer)