Solve Quadratics by Completing the Perfect Squares - Algebra-1

1. Fundamental Concepts

Completing the square is a method used to solve quadratic equations of the form \(ax^2 + bx + c = 0\) (\(a \neq 0\)) by transforming them into a perfect square form \((x + m)^2 = n\), which is then solved by taking the square root. Its core idea is to restructure the terms containing the unknown into a perfect square trinomial (in the form \(x^2 + 2mx + m^2 = (x + m)^2\)), simplifying the equation to enable solving via direct square root extraction.

A key feature of a perfect square trinomial is that the coefficient of the middle term is twice the product of the square roots of the first and last terms (e.g., in \(x^2 + 6x + 9 = (x + 3)^2\), \(6x = 2 \times x \times 3\) and \(9 = 3^2\)).

Equation Form Requirement: The coefficient of the quadratic term is usually converted to 1 first (if \(a \neq 1\), divide both sides by a) to facilitate constructing the perfect square.
Critical Step in Completing the Square: Add "the square of half the coefficient of the linear term" to both sides of the equation, turning the left side into a perfect square trinomial.
Connection with Square Root Extraction: After transforming into \((x + m)^2 = n\), if \(n \geq 0\), the equation has real solutions (\(x = -m \pm \sqrt{n}\)); if n < 0, the equation has no real solutions (since squares are non-negative).
Universality: Completing the square works for all quadratic equations, regardless of whether they are factorable, and serves as the basis for deriving the quadratic formula.

Equation: \(x^2 + 6x + 5 = 0\)

Steps:

Transpose the constant term to the right side: \(x^2 + 6x = -5\).
Complete the square: The coefficient of the linear term is 6, half of it is 3, and its square is 9. Add 9 to both sides:\(x^2 + 6x + 9 = -5 + 9\), which simplifies to \((x + 3)^2 = 4\).
Take the square root: \(x + 3 = \pm 2\).
Solve for x: \(x = -3 + 2 = -1\) or \(x = -3 - 2 = -5\).

Equation: \(2x^2 - 8x + 6 = 0\)

Steps:

Convert the coefficient of the quadratic term to 1: Divide both sides by 2, resulting in \(x^2 - 4x + 3 = 0\).
Transpose: \(x^2 - 4x = -3\).
Complete the square: The coefficient of the linear term is -4, half of it is -2, and its square is 4. Add 4 to both sides:\(x^2 - 4x + 4 = -3 + 4\), which simplifies to \((x - 2)^2 = 1\).
Take the square root: \(x - 2 = \pm 1\).
Solve for x: \(x = 2 + 1 = 3\) or \(x = 2 - 1 = 1\).

Equation: \(-3x^2 + 12x + 20 = 0\)

Steps:

Convert the coefficient of the quadratic term to 1: Divide both sides by -3, getting \(x^2 - 4x - \frac{20}{3} = 0\).
Transpose: \(x^2 - 4x = \frac{20}{3}\).
Complete the square: The coefficient of the linear term is -4, half of it is -2, and its square is 4. Add 4 to both sides:\(x^2 - 4x + 4 = \frac{20}{3} + 4\), which simplifies to \((x - 2)^2 = \frac{32}{3}\).
Take the square root: \(x - 2 = \pm \sqrt{\frac{32}{3}} \approx \pm 3.27\) (rounded to two decimal places).
Solve for x: \(x \approx 2 + 3.27 = 5.27\) or \(x \approx 2 - 3.27 = -1.27\).

Arrange the Equation Form: Rewrite the equation as \(ax^2 + bx + c = 0\), ensuring the coefficient of the quadratic term \(a \neq 0\).
Convert the Quadratic Coefficient to 1: If \(a \neq 1\), divide both sides by a to obtain \(x^2 + px + q = 0\) (where \(p = \frac{b}{a}\) and \(q = \frac{c}{a}\)).
Transpose Terms: Move the constant term to the right side of the equation, i.e., \(x^2 + px = -q\).
Complete the Square: Add "the square of half the coefficient of the linear term" (i.e., \((\frac{p}{2})^2\)) to both sides of the equation, turning the left side into a perfect square:\(x^2 + px + (\frac{p}{2})^2 = -q + (\frac{p}{2})^2\), which simplifies to \((x + \frac{p}{2})^2 = \frac{p^2}{4} - q\).
Solve by Taking Square Roots: If the result on the right side is non-negative (\(\frac{p^2}{4} - q \geq 0\)), take the square root of both sides to get \(x + \frac{p}{2} = \pm \sqrt{\frac{p^2}{4} - q}\), then solve for x; if the right side is negative, the equation has no real solutions.
Simplify the Result: If the solution is a radical, simplify it to its simplest form (e.g., \(\sqrt{\frac{32}{3}} = \frac{4\sqrt{6}}{3}\)); if it is a decimal, round it to the required number of decimal places.