Question

solve the quadratic equation by completing the square.

\\x^2 - 6x = -4\\

choose the appropriate process for rearranging the equation, if needed, before completing the square to obtain an equation of the form \\((x + k)^2 = d\\).

(type an integer or a simplified fraction.)

a. \\(x^2 - 6x = -4 \
ightarrow \frac{1}{6}x^2 - x + \box = -\frac{2}{3}\\)

b. \\(x^2 - 6x = -4 \
ightarrow x^2 - 6x + 9 = -4 + 9\\)

c. \\(x^2 - 6x = -4 \
ightarrow x^2 - 6x + \box = -4\\)

d. \\(x^2 - 6x = -4 \
ightarrow \frac{1}{6}x^2 - x + \box = -\frac{2}{3} + \box\\)

the solution set is \\(\\{\box\\}\\).
(type an exact answer, using radicals as needed. use a comma to separate answers as needed.)

Explanation:

Identify the correct process

To complete the square for \(x^2 - 6x = -4\), we find the constant to add.
Using Completing the Square:
Take half of the \(x\)-coefficient:

$$\frac{-6}{2} = -3$$

Square this value:

$$(-3)^2 = 9$$

Add \(9\) to both sides of the equation:

$$x^2 - 6x + 9 = -4 + 9$$

This matches option B.

Factor the perfect square trinomial

Using Perfect Square Trinomials:
Factor the left side of the equation:

$$(x - 3)^2 = 5$$

Solve for x

Take the square root of both sides:

$$x - 3 = \pm\sqrt{5}$$

Add \(3\) to both sides:

$$x = 3 \pm \sqrt{5}$$

The solution set is \(3 - \sqrt{5}, 3 + \sqrt{5}\).

Answer:

Question 1

• (A) \(x^2 - 6x = -4 \Rightarrow \frac{1}{6}x^2 - x + \Box = -\frac{2}{3}\)
• (B) \(x^2 - 6x = -4 \Rightarrow x^2 - 6x + 9 = -4 + 9\) (Correct answer)
• (C) \(x^2 - 6x = -4 \Rightarrow x^2 - 6x + \Box = -4\)
• (D) \(x^2 - 6x = -4 \Rightarrow \frac{1}{6}x^2 - x + \Box = -\frac{2}{3} + \Box\)

Question 2

The solution set is <blank>\(3 - \sqrt{5}, 3 + \sqrt{5}\)</blank>.