Algebra - Solving quadratic equations by completing the square-extended

Solve $x^2 + 6x - 4 = 0$ by completing the square. Leave your answer in simplest radical form.

$x^2 + 6x = 4$ $x^2 + 6x + (\frac{6}{2})^2 = 4 + (\frac{6}{2})^2$ $x^2 + 6x + 9 = 4 + 9$ $(x + 3)^2 = 13$ $x + 3 = \pm \sqrt{13}$ $x = -3 \pm \sqrt{13}$

First, move the constant $-4$ to the right side. Identify $b = 6$, calculate $(\frac{6}{2})^2 = 9$, and add it to both sides to create a perfect square trinomial on the left. Factor the left side and take the square root of both sides, remembering the plus-minus sign.

Solve $2x^2 - 8x + 3 = 0$ by completing the square.

$2(x^2 - 4x) + 3 = 0$ $2(x^2 - 4x + (-2)^2 - (-2)^2) + 3 = 0$ $2((x - 2)^2 - 4) + 3 = 0$ $2(x - 2)^2 - 8 + 3 = 0$ $2(x - 2)^2 - 5 = 0$ $2(x - 2)^2 = 5$ $(x - 2)^2 = \frac{5}{2}$ $x - 2 = \pm \sqrt{\frac{5}{2}}$ $x = 2 \pm \frac{\sqrt{10}}{2}$

Since $a = 2$, factor $2$ out of the first two terms. Complete the square inside the parentheses by adding and subtracting $(\frac{-4}{2})^2 = 4$. Distribute the $2$ back to the $-4$ constant, simplify, and solve for $x$ by isolating the squared term and taking the square root.

Solve $x^2 + 5x + 2 = 0$ by completing the square.

$x^2 + 5x = -2$ $x^2 + 5x + (\frac{5}{2})^2 = -2 + (\frac{5}{2})^2$ $x^2 + 5x + \frac{25}{4} = -\frac{8}{4} + \frac{25}{4}$ $(x + \frac{5}{2})^2 = \frac{17}{4}$ $x + \frac{5}{2} = \pm \frac{\sqrt{17}}{2}$ $x = -\frac{5}{2} \pm \frac{\sqrt{17}}{2}$ $x = \frac{-5 \pm \sqrt{17}}{2}$

Identify $b = 5$. Add $(\frac{5}{2})^2 = \frac{25}{4}$ to both sides. Convert the constant $-2$ to a fraction with a common denominator ($-\frac{8}{4}$). Factor into a square of a binomial, take the square root of both sides, and solve for $x$.