Question

solve \\(x^2 = 12x - 15\\) by completing the square. which is the solution set of the equation?

\\(\\{-6 - \sqrt{51}, -6 + \sqrt{51}\\}\\)
\\(\\{-6 - \sqrt{21}, -6 + \sqrt{21}\\}\\)
\\(\\{6 - \sqrt{51}, 6 + \sqrt{51}\\}\\)
\\(\\{6 - \sqrt{21}, 6 + \sqrt{21}\\}\\)

Explanation:

Rearrange the quadratic equation

We start with the given equation:

$$x^2 = 12x - 15$$

Using the Quadratic Equations knowledge point, we move the linear term to the left side:

$$x^2 - 12x = -15$$

Complete the square

To complete the square, we find half of the coefficient of \(x\), square it, and add it to both sides.
The coefficient of \(x\) is \(-12\). Half of \(-12\) is \(-6\), and \((-6)^2 = 36\).
Add \(36\) to both sides:

$$x^2 - 12x + 36 = -15 + 36$$

$$(x - 6)^2 = 21$$

Solve for x

Using the Quadratic Roots knowledge point, we take the square root of both sides:

$$x - 6 = \pm\sqrt{21}$$

Add \(6\) to both sides to isolate \(x\):

$$x = 6 \pm\sqrt{21}$$

Write the solution set

The solutions are:

$$x = 6 - \sqrt{21} \quad \text{and} \quad x = 6 + \sqrt{21}$$

Thus, the solution set is:

$$\{6 - \sqrt{21}, 6 + \sqrt{21}\}$$

Answer:

• (A) \(\{-6 - \sqrt{51}, -6 + \sqrt{51}\}\)
• (B) \(\{-6 - \sqrt{21}, -6 + \sqrt{21}\}\)
• (C) \(\{6 - \sqrt{51}, 6 + \sqrt{51}\}\)
• (D) \(\{6 - \sqrt{21}, 6 + \sqrt{21}\}\) (Correct answer)