Question

question 5 (5 points)
find the solutions of the quadratic equation \\(x^2 - 6x + 8 = 0\\).
a) \\(x = -2, 4\\)
b) \\(x = -2, -4\\)
c) \\(x = 2, 4\\)
d) \\(x = 2, -4\\)

question 6 (5 points)
what is the solution to the inequality \\(x^2 > -9x - 18\\)?
a) \\(x < -6\\) or \\(x > -3\\)
b) \\(-6 < x < -3\\)
c) \\(x \le -6\\) and \\(x \ge -3\\)
d) \\(x < -6\\) and \\(x > -3\\)

question 7 (5 points)
apply the square root principle to solve \\((x - 2)^2 + 20 = 0\\).
a) \\(x = -2 + 2i\sqrt{5}, x = -2 - 2i\sqrt{5}\\)
b) \\(x = 2 + 2i\sqrt{5}, x = 2 - 2i\sqrt{5}\\)
c) \\(x = -2 + 2i\sqrt{5}, x = -2 - 2i\sqrt{5}\\)
d) \\(x = 2 + 2\sqrt{5}, x = 2 - 2\sqrt{5}\\)

Explanation:

Solve Question 5

Using the Quadratic Formula knowledge point

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Solve Question 6

Using the Quadratic Inequalities knowledge point

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Isolate the squared term

We begin solving Question 7 by isolating the squared binomial on one side of the equation.

$$ (x - 2)^2 = -20 $$

Apply the square root principle

We take the square root of both sides, introducing a imaginary unit \(i\) since we are taking the square root of a negative number.

$$ x - 2 = \pm \sqrt{-20} $$

$$ x - 2 = \pm i\sqrt{20} $$

Simplify the radical and solve

We simplify the radical \(\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}\) and solve for \(x\).

$$ x - 2 = \pm 2i\sqrt{5} $$

$$ x = 2 \pm 2i\sqrt{5} $$

This gives the two solutions:

$$ x = 2 + 2i\sqrt{5}, \quad x = 2 - 2i\sqrt{5} $$

Answer:

Question 5

• (A) \(x = -2, 4\)
• (B) \(x = -2, -4\)
• (C) \(x = 2, 4\) (Correct answer)
• (D) \(x = 2, -4\)

Question 6

• (A) \(x < -6\) or \(x > -3\) (Correct answer)
• (B) \(-6 < x < -3\)
• (C) \(x \le -6\) and \(x \ge -3\)
• (D) \(x < -6\) and \(x > -3\)

Question 7

• (A) \(x = -2 + 2\sqrt{5}, x = -2 - 2\sqrt{5}\)
• (B) \(x = 2 + 2i\sqrt{5}, x = 2 - 2i\sqrt{5}\) (Correct answer)
• (C) \(x = -2 + 2i\sqrt{5}, x = -2 - 2i\sqrt{5}\)
• (D) \(x = 2 + 2\sqrt{5}, x = 2 - 2\sqrt{5}\)