Problem 42 - Introductory Algebra for College Students

Question

Explain how to solve \(x^{2}+6 x+8=0\) by completing the square.

Step-by-Step Solution

Answer

The solution to the equation \(x^{2}+6 x+8=0\) by completing the square method is \(x = -3 ± \sqrt{17}\)

Step 1: Rearrange the equation and identify a and b

Rearrange the equation in the form \(x^{2} + bx + c = 0\). Here in the given equation, \(x^{2} + 6x + 8 = 0\), a is 1, b is 6, and c is 8.

Step 2: Complete the Square

To make it a perfect square trinomial, we'll subtract c from both sides and add \((\frac{b}{2a})^{2}\) to both sides. This gives us: \(x^{2}+6 x = 8\), then \(x^{2}+6 x + (\frac{6}{2*1})^{2} =8+ (\frac{6}{2*1})^{2}\). It simplifies to: \(x^{2}+6 x + 9 = 17\)

Step 3: Write in Square Form and Solve

The left-hand side of the equation is now a perfect square: \((x+3)^{2}=17\). Now, get the square root on both sides: \(x+3 = ± \sqrt{17}\). Subtract 3 from both sides to solve for x: \(x = -3 ± \sqrt{17}\)