Problem 158 - College Algebra Essentials

Question

Explain how to solve \(x^{2}+6 x+8=0\) using the quadratic formula.

Step-by-Step Solution

Answer

The solutions to the quadratic equation \(x^{2}+6 x+8=0\) are \(x=-2\) and \(x=-4\).

Step 1: Determine the Coefficients

Identify the coefficients in the standard form \(ax^2 + bx + c = 0\). In the equation \(x^2 + 6x + 8=0\), \(a=1\), \(b=6\), and \(c=8\).

Step 2: Substitute the values into Quadratic Formula

Substitute these coefficients \(a\), \(b\), and \(c\) into the quadratic formula \(x= -b \pm \sqrt{b^2-4ac} / 2a\). This will result in \(x= -6 \pm \sqrt{6^2-4(1)(8)} / 2(1)\).

Step 3: Simplify under the square root

Calculate the expression under the square root, \(b^2-4ac\), which becomes \(36-32\). This simplifies to \(\sqrt{4}\). Hence, the equation will now look like \(x= -6 \pm \sqrt{4} / 2\).

Step 4: Continue simplification

Simplify \(\sqrt{4}\) to \(2\). The equation will now look like \(x= -6 \pm 2 / 2\).

Step 5: Calculate The Roots

Next, solve for \(x\) for the plus and minus cases separately. The roots will be \(x= (-6+2)/2=-2\) and \(x= (-6-2)/2=-4\).